Kirchhoff's Laws and Electrical Energy

KIRCHHOFF'S LAWS

• Kirchhoff's laws are used to solve complicated electric circuits that cannot be reduced to simple series-parallel combinations.

• These laws are based on the conservation of electric charge and energy.

• Examples of such circuits are shown in Fig. 3.20 (not included here).

• Kirchhoff's laws were derived in 1845 and 1846.

Junction

• A junction is a point in a circuit where three or more conductors meet.

• Junctions are also called nodes or branch points.

• In Fig. 3.20 (a), points D and C are junctions, and in Fig. 3.20 (b), points B and F are junctions.

Loop

• A loop is any closed conducting path.

• In Fig. 3.20 (a), examples of loops are ABCDA, DCEFD, and ABEFA. In Fig. 3.20 (b), examples are CBFEC and BDGFB.

Kirchhoff's Junction Rule (KCL)

• The algebraic sum of the currents meeting at a junction in an electric circuit is always zero.

• Mathematically, this is represented as: $\Sigma i = 0$

  • where the summation is over all currents at the junction.

• Alternatively, the sum of currents directed towards a node equals the sum of currents directed away from the node.

• Also known as Kirchhoff's Current Law (KCL).

• The junction rule is based on the conservation of electric charge.

• In Fig. 3.21 (not provided), according to KCL: $I<em>1 + I</em>2 = I<em>3 + I</em>4$

Kirchhoff's Loop Rule (KVL)

• The algebraic sum of the potential changes around any closed loop involving resistors and cells is zero.

• In any closed loop of an electrical circuit, the algebraic sum of the electromotive forces (EMFs) is equal to the algebraic sum of the products of the resistances and the currents flowing through them.

• Mathematically, this is represented as: $\Sigma \Delta V = 0$

  • sum over the closed loop

• Also known as Kirchhoff's Voltage Law (KVL) or loop rule.

• The loop rule is based on the conservation of energy.

Sign Conventions for Applying Kirchhoff's Laws

1. Potential Change Across a Resistance:

   • When traversing a resistance in the direction of the current, the change in potential is $-IR$.

   • When traversing a resistance in the opposite direction of the current, the change in potential is $+IR$.

2. Potential Change Across an EMF Source:

   • When traversing an EMF source from the negative to the positive terminal, the change in potential is $+E$.

   • When traversing an EMF source from the positive to the negative terminal, the change in potential is $-E$.

   • This is irrespective of the direction of the current in the circuit.

Calculating Potential Difference Using Kirchhoff's Laws

1. Start from a point on the loop and traverse the loop (either clockwise or counterclockwise) to return to the same point.

2. Balance currents at each junction using KCL.

3. If moving along the direction of current, there is a potential drop across the resistance; moving in the opposite direction results in a potential gain.

4. The net sum of all potential differences should be zero, according to KVL.

• Example Circuit (Fig. 3.24, not provided):

  • In closed loop ABEFA: $-E<em>1 - i</em>2 R<em>3 - iR</em>2 - iR_1 = 0$

  • In closed loop BCDEB: $E<em>2 + i</em>2 R<em>3 + E</em>3 - (i<em>1 + i</em>2) R_4 = 0$

Example 3.43: Circuit Analysis

• Problem: Find currents in different branches of the electric circuit shown. Find the potential difference between points A and B. Find the current through the 20V cell if points A and B are connected.

• Solution:

  • Applying Kirchhoff's first law (junction law) at junction B: $i<em>1 = i</em>2 + i_3$

  • Applying Kirchhoff's second law in loop 1 (ABEFA): $-4 + 4i<em>1 + 2i</em>2 = 0$

  • Applying Kirchhoff's second law in loop 2 (BCDEB): $-2i<em>3 - 6 - 4i</em>3 - 4 = 0$

  • Solving these equations:

    • $i_1 = 1 A$

    • $i_2 = \frac{8}{3} A$

    • $i_3 = -\frac{5}{3} A$

    • The negative sign of $i_3$ indicates that the actual current direction is opposite to what was assumed.

  • Potential difference between A and B:

    • $V<em>A - 9 - 3 \times 1 - 8 - 3 \times 2 - 7 = V</em>B$

    • $V<em>A - V</em>B = 33 V$

  • With A and B connected, resistance $2\Omega$, $2\Omega$ and $1\Omega$ are in series along APSC, and resistances $1\Omega$, $2\Omega$ and $2\Omega$ are in series along AQRC.

  • No current flows in branch CB as AB is not connected.

  • Let the current in circuit APSCRQA be $i$. Using KVL,

  • $V<em>A - i \times 2 - 20 - i \times 2 - i \times 1 - i \times 2 - i \times 2 + 10 - i \times 1 = V</em>A$

  • $10 = -10i \implies i = -1A$

  • $V<em>A - V</em>B = 15 + 5i = 15 + 5 \times (-1) = 10V$

Example 3.44: Potential Difference Calculation

• Problem: Find the potential difference between points F and C in the previous example.

• Solution: Reaching from F to C via A and B:

  • $V<em>F + 2 + 4i</em>1 - 2i<em>3 = V</em>C$

  • $V<em>F - V</em>C = 4i<em>1 + 2i</em>3 - 2$

  • Substituting $i<em>1 = 1A$ and $i</em>3 = -\frac{5}{3}A$, we get:

  • $V<em>F - V</em>C = \frac{4}{3}V$

    • The negative sign implies VF < VC.

Example 3.45: Potential Difference with Multiple Batteries

• (i) Problem: Find the potential difference between points A and B given multiple voltage sources and resistors.

• Solution:

  • Resistance $2\Omega$, $2\Omega$, and $1\Omega$ are in series along APSC. Similarly, $1\Omega$, $2\Omega$, and $2\Omega$ are in series along AQRC.

• The equivalent voltage is:

  • $E = \frac{\frac{20}{5} + \frac{5}{5} + \frac{10}{5}}{\frac{1}{5} + \frac{1}{5} + \frac{1}{5}} = \frac{\frac{7}{1}}{\frac{3}{5}} = \frac{35} {3}$

  • $E = \frac{\frac{20}{5} + \frac{5}{5} + \frac{10}{5}}{\frac{1}{5} + \frac{1}{5} + \frac{1}{5}} = \frac{4+1+2}{\frac{1}{5} + \frac{1}{5} + \frac{1}{5}} = \frac{7}{\frac{3}{5}} = \frac{35} {3}$

• Potential difference between points A and B:

  • $V<em>A - V</em>B = \frac{55}{7} V$

• For the cell of EMF 20V:

  • $V<em>A - V</em>B = 20 - 5i_1$

  • $i_1 = \frac{12}{7} A \approx 2.4 A$

Electrical Energy

• Electrical energy is the total work done (W) by the source of EMF in maintaining an electric current (I) in the circuit for a specified time (t).

• $W = qV$

• Using Ohm's law ($V = IR$), total charge that crosses the resistor is given by $q = It$

• Energy gained is: $E = W = Vq = VIt = I^2Rt = \frac{V^2}{R}t \text{ (joule)}$

• In calories: $E = \frac{VIt}{J} = \frac{I^2Rt}{J} = \frac{V^2t}{RJ} \text{ (cal)}$ where J is the mechanical equivalent of heat (4.186 J/cal).

• These relations are also called Joule's law of heating.

• The SI unit of electrical energy is the joule (J).

Electrical Power

• Electrical power is the rate of electrical energy supplied per unit time to maintain the flow of electric current through a conductor.

• Mathematically, $P = \frac{W}{t} = VI = I^2R = \frac{V^2}{R}$

• The SI unit of power is the watt (W), where 1 watt = 1 volt × 1 ampere.

• Bigger units of electrical power are kilowatt (kW) and megawatt (MW), where 1 kW = 1000 W and 1 MW = $10^6$ W.

• Commercial unit of electrical power is horsepower (HP), where 1 HP = 746 W.

Heating Effects of Current

• The electric current through a resistor increases its thermal energy.

• This is called the heating effect of electric current.

• When some potential difference V is applied across a resistance R, charge q flows through the circuit in time t, and the heat absorbed or produced is given by:

Example 3.46: Heat Developed Across Resistors

• Problem: Find the heat developed across each resistance in 2 seconds in a given network of resistors.

• Solution:

  • The 6Ω and 3Ω resistances are in parallel, so their combined resistance is:

  • $\frac{1}{R} = \frac{1}{6} + \frac{1}{3} \implies R = 2\Omega$

  • The equivalent simple circuit can be drawn as shown.

• Current in the circuit:

  • $i = \frac{\text{Net emf}}{\text{Total resistance}} = \frac{20}{3 + 2 + 5} = 2A$

• Potential difference across the 6Ω and 3Ω resistances is the same:

  • $V = iR = 2 \times 2 = 4V$

• Heat developed across the resistors:

  • $H_3 = \frac{V^2}{R} t = \frac{4^2}{3} \times 2 = \frac{32}{3} J$

  • $H_6 = \frac{V^2}{R} t = \frac{4^2}{6} \times 2 = \frac{16}{3} J$

  • $H_5 = I^2 R t = 2^2 \times 5 \times 2 = 40 J$

Electricity Consumption

• A special unit, the kilowatt-hour (kWh), is used commercially to measure electrical energy consumed.

• Also called 1 unit of electrical energy.

• 1 kilowatt-hour (kWh) is the amount of energy dissipated in 1 hour in a circuit when the electric power in the circuit is 1 kilowatt.

• 1 kilowatt hour (kWh) = $3.6 \times 10^6$ joules (J)

*Note: Resistance of electrical appliance On electrical appliances (bulbs, geysers, heaters, etc.) wattage, voltage printed are called rated values. The resistance of any electrical appliance can be calculated by rated power and rated voltage by using $R = \frac{V<em>R^2}{P</em>R}$.

Power Consumption in Combinations of Bulbs

Series Combination of Bulbs

• (i) Total power consumed is given by:

  • $\frac{1}{P<em>{eq}} = \frac{1}{P</em>1} + \frac{1}{P_2} + …$

• (ii) Power consumed (brightness) is proportional to Voltage, Resistance, and inversely proportional to the rated power

  • $P<em>{\text{consumed}} \propto V \propto R \propto \frac{1}{P</em>{\text{rated}}}$.
    Combinations bulb of lesser wattage will give more bright light and potential difference appearing across it will be more.

Parallel Combination of Bulbs

• (i) Total power consumed is given by:

  • $P<em>{\text{total}} = P</em>1 + P_2 + …$

• (ii) Power consumed (brightness) is:

  • $P<em>{\text{consumed}} \propto P</em>{\text{rated}}$
    Combination, bulb of greater wattage will give more bright light and more current will pass through it.

Applications of Heating Effects of Current

• (i) Filament of electric bulb:

  • Made of tungsten, which has high resistivity and high melting point.

• (ii) Electric devices having heating elements:

  • Like heaters, geysers, or presses, are made of nichrome, which has high resistivity and high melting point.

• (iii) Fuse wire:

  • Made of tin-lead alloy, which has a low melting point and a high resistivity.

  • Used in series as a safety device in an electric circuit, designed to melt and thereby open the circuit if the current exceeds a predetermined value due to some fault.

Example 3.49: Brightness of Bulbs in Series and Parallel

• Problem: Two bulbs, one rated 60 W-220 V and the other 100 W-220 V, are joined (i) in series and (ii) in parallel. Which bulb will glow brighter in each case?

• Solution:

  • Parallel Combination: In case of parallel combination, same voltage will appear on each bulb, hence heat produced will be proportional to (1/R). As bulb of greater wattage will glow more. Hence, 100 W bulb will glow brighter.

  • Series Combination: In case of series combination, same current flows through each bulb, hence heat produced will be proportional to R (P=I^2R). As we know, higher the wattage, lower the resistance, then 60 W bulb will have higher resistance. That means more heat will be produced in 60 W bulb in this case, so this bulb will glow brighter.

Example 3.50: Fusing of Bulbs in Series

• Problem: Two bulbs rated 40 W-220 V and 100 W-220 V are connected in series, and alternately, (i) 300 V and (ii) 440 V is applied. Determine which bulb will fuse in each case.

• Solution:

  • 1. Find maximum current each bulb can bear:

  • For the 40W bulb:

  • $P = VI = 40 = 220I \Rightarrow I = \frac{40}{220} \approx 0.18A$

  • For the 100W bulb:

  • $I = \frac{100}{220} \approx 0.45A$

  • Calculate the resistance of each bulb (using $R = \frac{V^2}{P}$):

  • $R_{40} = \frac{220^2}{40} = 1210\Omega$

  • $R_{100} = \frac{220^2}{100} = 484\Omega$

  • (i) With 300V applied:

  • Total resistance = $1694\Omega$. Current in each bulb: $I = \frac{300}{1694} \approx 0.177A$.

  • For (i), the current (0.177A), which is less than their maximum permissible current. So no bulb will fuse

  • (ii) With 440V applied:

  • $I = \frac{440}{1694} \approx 0.26A$. This greater than 0.18A current but less than 0.45A

  • For (ii), the 40 W bulb will fuse.

Example 3.51: Fusing of Bulbs in Parallel

• Problem: Two bulbs from above example are joined in parallel configuration from the previous example what would happen? and 300 V is applied, which of the two bulbs will fuse?

• Solution If they are joined in parallel and 300 V is applied on them, then both will get 300 V. Since, their rating is 220 V, naturally, current flowing through them will be more than maximum possible value. Hence, both will fuse out.

Example 3.52: Boiling Water with Bulbs

• Problem: Two electric heaters rated 60 W-220 V and 100 W-220 V boiling certain amount of water. If they are operating at 220 V takes time 2 min and 1.5 min separately and (i) in series and (ii) in parallel, then find the ratio of time taken by the two cases to boil the same water.

• Solution: Given the heaters are rated 60 W and 100 W at 220 V.

  • In series, the total power output would be:

  • $P = \frac{P<em>1P</em>2}{P<em>1 + P</em>2} = \frac{60 \times 100}{60 + 100} = 37.5 W$

  • In parallel, the total power output would be:

  • $P = P<em>1 + P</em>2 = 60 + 100 = 160 W$

• In series, the equivalent resistance would be:

• Times taken to boil water is inversely proportional to power, the ratio of time is:

• $160/37.5 = 4.27$ or 4.3

Example 3.53: Illumination and Power Dissipation

• Figures shows three identical bulbs A, B and C

• Problem: Figure shows three identical bulbs A, B and C, which are connected to a battery of supply voltage V. When the switch S is closed, then discuss the change in (i) the illumination of the three bulbs. (ii) the power dissipated in the circuit.

• Solution:

  • When the switch S is open, the bulbs will be in series with one another.

  • (i) When the switch S is closed, then the bulb C is short circuited and hence there will be no current through C

  • $\implies, P = 0$
    Therefore, the intensity of illumination of each of the bulb A and B become 9/4 times of the initial value but
    the intensity of the bulb C becomes zero.
    (i) The power dissipated in the circuit before closing the
    switch is

Check Point 3.3

Some questions answered below

• Question 2: In the circuit element given here, if the potential difference at point B, $V_B = 0$, then the potential difference between A and D are

  • Answer: (d)

• Question 4: How much work is required to carry a 6μC charge from the negative terminal to the positive terminal of a 9 V battery?

  • Answer: (c)

• Question 5: Two resistors R and 2R are connected in series in an electric circuit. The thermal energy developed in R and 2R are in the ratio

  • Answer: (a)

• Question 11: If in the circuit, power dissipation is 150 W, then R is

  • Answer: (d)

• Question 15: A and B are two bulbs connected in parallel. If A is glowing brighter than B, then the relation between RA and Rg is

  • Answer: (b)