Question

When a current of 0.5 A is passed through
two resistances in series, the potential difference between the
ends of the series arrangement is 12.5 V. On connecting
them in parallel and passing a current of 1.5 A, the potential
difference between their ends is 6 V. Calculate the two
resistances.​

Answer 1

We will solve question in two cases, as given below:

Case 1

• Voltage = 12.5 Volts
• i = 0.5 Amphere

→ R = V/i

→ R = 12.5 V/0.5 A

→ R = 125/5 Ω

→ R = 25 Ω

Hence in series combination, resistance is 25 Ω.

Case 2

• Voltage = 6 Volts
• i = 1.5 Amphere

→ R = V/i

→ R = 6 V/1.5 A

→ R = 60/15 Ω

→ R = 4 Ω

Hence resistance in parallel combination is 4 Ω.

Answer 2

Answer:

current=0.5A , V=12.5V

in series Resistance= 12.5/0.5 = 25

In parallel current=1.5A ,Potential difference= 6V

resistance=6/1.5 = 4