Electricity Past Paper PDF 2024-25

n Take out the plug from the key. Remove the ammeter and voltmeter from the circuit. Insert the ammeter in series with the resistor R1, as shown in Fig. 11.11. Note the ammeter reading, I1. Figure 11.11 n Similarly, measure...

n Take out the plug from the key. Remove the ammeter and voltmeter from the circuit. Insert the ammeter in series with the resistor R1, as shown in Fig. 11.11. Note the ammeter reading, I1. Figure 11.11 n Similarly, measure the currents through R 2 and R 3. Let these be I 2 and I 3, respectively. What is the relationship between I, I1, I2 and I3 ? It is observed that the total current I, is equal to the sum of the separate currents through each branch of the combination. I = I1 + I2 + I3 (11.15) Let Rp be the equivalent resistance of the parallel combination of resistors. By applying Ohm’s law to the parallel combination of resistors, we have I = V/Rp (11.16) On applying Ohm’s law to each resistor, we have I1 = V /R1; I2 = V /R2; and I3 = V /R3 (11.17) From Eqs. (11.15) to (11.17), we have V/Rp = V/R1 + V/R2 + V/R3 or 1/Rp = 1/R1 + 1/R2 + 1/R3 (11.18) Thus, we may conclude that the reciprocal of the equivalent resistance of a group of resistances joined in parallel is equal to the sum of the reciprocals of the individual resistances. Example 11.8 In the circuit diagram given in Fig. 11.10, suppose the resistors R1, R2 and R 3 have the values 5 Ω, 10 Ω, 30 Ω, respectively, which have been connected to a battery of 12 V. Calculate (a) the current through each resistor, (b) the total current in the circuit, and (c) the total circuit resistance. Solution R1 = 5 Ω, R2 = 10 Ω, and R3 = 30 Ω. Potential difference across the battery, V = 12 V. This is also the potential difference across each of the individual resistor; therefore, to calculate the current in the resistors, we use Ohm’s law. The current I1, through R1 = V/ R1 I1 = 12 V/5 Ω = 2.4 A. 186 Science 2024-25