Question

The battery in a circuit has emf of 50V . The current in the lamp is 2A and the reading on the voltmeter is 30V.
Calculate the internal resistanse of the battery and the resistance of the lamp.

Answer 1

Answer on Question 62349, Physics, Electric Circuits

Question:

The battery in a circuit has emf of $50\,V$. The current in the lamp is $2\,A$ and the reading on the voltmeter is $30\,V$. Calculate the internal resistance of the battery and the resistance of the lamp.

Solution:

a) Let's denote the potential difference across the lamp as $V$ and the potential drop across the battery as $V_r$. Then, we can write the formula for the electromotive force of the battery:

$$\mathcal{E} = V + V_r.$$

We can find $V_r$ from the Ohm's law:

$$V_r = I r.$$

Let's substitute $V_r$ into the previous formula, we get:

$$V = \mathcal{E} - I r,$$

here, $V$ is the potential difference across the lamp, $\mathcal{E}$ is the electromotive force of the battery, $I$ is the current in the lamp, $r$ is the internal resistance of the battery.

From this formula we can find the internal resistance of the battery:

$$r = \frac{\mathcal{E} - V}{I} = \frac{50\,V - 30\,V}{2\,A} = 10\,\Omega.$$

b) We can find the resistance of the lamp from the Ohm's law:

$$R = \frac{V}{I} = \frac{30\,V}{2\,A} = 15\,\Omega.$$

Answer:

a) $r = 10\,\Omega$, b) $R = 15\,\Omega$.

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