Question

A battery of emf 12 V and internal resistance 2 Ω is connected with two resistors A and B of resistance 4 Ω and 6 Ω respectively joined in series.

Find:

1. Current in the circuit.
2. The terminal voltage of the cell.
3. The potential difference across 6Ω Resistor.
4. Electrical energy spent per minute in the 4Ω resistor.

A battery of emf 12 V and internal resistance 2 Ω is connected with two resistors A and B of resistance 4 Ω and 6 Ω respectively joined in series.

Find:

1. Current in the circuit.
2. The terminal voltage of the cell.
3. P.D across 6 Ω resistor.
4. Electrical energy spent per minute in the 4Ω resistor.

Answer 1

The external resistors and internal resistance are all in series:

\[R_{\mathrm{total}}=4+6+2=12\ \Omega\]

= 12 Ω

(i) Current:

\[I=\frac{12}{12}\]

= 1 A

(ii) Terminal voltage is the voltage across the external circuit:

\[V_{\mathrm{terminal}}=I(4+6)\]

= 1 × 10

= 10 V

(iii) Potential difference across the \(6\ \Omega\) resistor:

\[V_6=IR\]

= 1 × 6

= 6 V

(iv) Energy spent in the \(4\ \Omega\) resistor in one minute:

\[E=I^2Rt\]

\[=1^2\times4\times60\]

= 240 J