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The current in a simple series circuit is 5.0 amp. When anadditional resistance of 2.0 ohm is inserted, the current30.drops to 4.0 amp. The original resistance of the circuit inohm was[KCET 2005)

Answer»

I = 5 A Resistance = RI' = 4 A R' = R+2 Ω = total resistance

We assume R is the total effective resistance in the circuit in series with the battery. Also, that 2 Ω resistance is added in series with the= given resistance and battery.

Voltage of the battery = E = I R = I' R' 5 * R = 4 ( R +2) => R = 8 Ω battery potential is = 40 V

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