Question

A 10 m long wire of resistance 20 ohm is connected in series with a battery of emf 3 V and a resistance of 10 ohm. The potential gradient along with wire is

Answer 1

Answer:

The  voltage gradient along the wire is 0.2 v/m

Explanation:

Net resistance of the circuit = 10 + 20 = 30 Ω

Voltage of the battery = 3 V

=> current = voltage/resistance

= 3/30

= 0.1 A

Voltage drop along the line,

V = current x resistance

=> V = 0.1 x 20 = 2 V

Length of the line = 10 m

=> voltage gradient along the wire = voltage/length = 2/10 = 0.2 v/m

Hence the  voltage gradient along the wire is 0.2 v/m

Answer 2

Net resistance of the circuit = 10 + 20 = 30 Ω

Voltage of the battery = 3 V

=> current = voltage/resistance

= 3/30

= 0.1 A

Voltage drop along the line,

V = current x resistance

=> V = 0.1 x 20 = 2 V

Length of the line = 10 m

=> voltage gradient along the wire = voltage/length = 2/10 = 0.2 v/m

Hence the  voltage gradient along the wire is 0.2 v/m

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