Electricity PDF Example 11.9 - 2024

The current I2, through R 2 = V/ R2 I2 = 12 V/10 Ω = 1.2 A. The current I3, through R 3 = V/R3 I3 = 12 V/30 Ω = 0.4 A. The total current in the circuit, I = I1 + I2 + I3 = (2.4 + 1.2 + 0.4) A...

The current I2, through R 2 = V/ R2 I2 = 12 V/10 Ω = 1.2 A. The current I3, through R 3 = V/R3 I3 = 12 V/30 Ω = 0.4 A. The total current in the circuit, I = I1 + I2 + I3 = (2.4 + 1.2 + 0.4) A = 4A The total resistance Rp, is given by [Eq. (11.18)] 1 1 1 1 1 = + + = R p 5 10 30 3 Thus, Rp = 3 Ω. Example 11.9 If in Fig. 11.12, R 1 = 10 Ω, R2 = 40 Ω, R3 = 30 Ω, R4 = 20 Ω, R5 = 60 Ω, and a 12 V battery is connected to the arrangement. Calculate (a) the total resistance in the circuit, and (b) the total current flowing in the circuit. Solution Suppose we replace the parallel resistors R1 and R 2 by an equivalent resistor of resistance, R ′. Similarly we replace the parallel resistors R3, R4 and R5 by an equivalent single resistor of resistance R ″. Then using Eq. (11.18), we have 1/ R′ = 1/10 + 1/40 = 5/40; that is R′ = 8 Ω. Similarly, 1/ R″ = 1/30 + 1/20 + 1/60 = 6/60; that is, R″ = 10 Ω. Thus, the total resistance, R = R′ + R″ = 18 Ω. To calculate the current, we use Ohm’s law, and get I = V/R = 12 V/18 Ω = 0.67 A. We have seen that in a series circuit the current is constant throughout the electric circuit. Thus it is obviously impracticable to Figure 11.12 connect an electric bulb and an electric heater in series, because they An electric circuit showing need currents of widely different values to operate properly (see Example the combination of series and parallel resistors 11.3). Another major disadvantage of a series circuit is that when one component fails the circuit is broken and none of the components works. If you have used ‘fairy lights’ to decorate buildings on festivals, on marriage celebrations etc., you might have seen the electrician spending lot of time in trouble-locating and replacing the ‘dead’ bulb – each has to be tested to find which has fused or gone. On the other hand, a parallel circuit divides the current through the electrical gadgets. The total resistance in a parallel circuit is decreased as per Eq. (11.18). This is helpful particularly when each gadget has different resistance and requires different current to operate properly. Electricity 187 2024-25