Question

In the opposite electric circuit, if the reading of the

voltmeter is 6 V, the resistance of R equals

Answer 1

`2\; {\rm \Omega}`.

Step-by-step explanation:

The lower case "
`r = 1\; {\rm \Omega}`" in the circuit diagram indicates that the internal resistance of this
`12\; {\rm V}`-battery is
`1\; {\rm \Omega}`.

In practice, the rated
`12\; {\rm V}` voltage drop across this battery is split across not just the resistors in the external circuit, but also across the internal resistance of the battery.

The voltameter in this circuit measures the voltage drop across the
`3\; {\rm \Omega}` resistor and the unknown resistor
`R`. The voltage drop across the internal resistance and the other resistor (a total of
`1\; {\rm \Omega} + 4\; {\rm \Omega} = 5\; {\rm \Omega}`) would be
`12\; {\rm V} - 6\; {\rm V} = 6\; {\rm V}`.

Divide the voltage drop by the resistance to find the current:

`\begin{aligned}I &= (V)/(R) \\ &= \frac{6\; {\rm V}}{5\; {\rm \Omega}} = 1.2\; {\rm A}\end{aligned}`.

Since this circuit is serial, current would be the same everywhere in the circuit. Given that voltage drop across the
`3\; {\rm \Omega}` resistor and the unknown resistor is
`6\; {\rm V}`, the resistance of that part of the circuit would be:

`\begin{aligned}\frac{6\; {\rm V}}{1.2\; {\rm A}} = 5\; {\rm \Omega}\end{aligned}`.

Subtract the
`3\; {\rm \Omega}` from the resistance to find the value of
`R`:

`R = (5\; {\rm \Omega} - 3\; {\rm \Omega}) = 2\; {\rm \Omega}`.