Question

You are measuring the current in a circuit that is operated on an 18v battery. The ammeter reads 40mA. Later you notice the current has dropped to 20mA. How much has the voltage changed.

Answer 1

Answer:

answer is 9V .........

Answer 2

Answer:

The voltage changed, $V_{2}$, after the current dropped measured is $9V$.

Explanation:

Given,

The initial voltage of the circuit, $V_{1}$ = $18V$

The initial current flow in the circuit, $I_{1}$ = $40mA$ = $0.04A$

Later, the current dropped, $I_{2}$ = $20mA$ = $0.02A$

The voltage changed after the current dropped, $V_{2}$ =?

As we know,

• Voltage = current × resistance

Therefore,

• $V_{1} = I_{1}\times R$
• Resistance, R = $\frac{V_{1} }{I_{1} }$ = $\frac{18}{0.04}$ = $450\Omega$

After the current dropped, $I_{2}$ = $0.02A$ and, as calculated, R = $450\Omega$.

Therefore,

• $V_{2} = I_{2}\times R$
• $V_{2} = 0.02\times 450$ = $9V$.

Hence, the voltage changed after the current dropped, $V_{2}$ = $9V$.