Question

A closed box has two metal terminals a and b. The inside of the box contains an unknown emf 8 in series with a resistance R. When a potential difference of 22.0 V is maintained between terminal a and terminal b, there is a current of 1.50 A between the terminals a and b. If this potential difference is reversed, a current of 1.90 A in the reverse direction is observed. Find epsilon and R.

Answer 1

Answer:

The value of $\epsilon$ and R are 12.9 ohm and 2.6 V.

Explanation:

Given that,

Potential difference = 22.0 V

Current = 1.50 A

If this potential difference is reversed,

Then the current = 1.90 A

In forward direction

The net emf is

$\epsilon=(\epsilon+22.0)\ V$

The net current is

$I=\dfrac{V}{R}$

$1.50=\dfrac{\epsilon+22.0}{R}$...(I)

In reverse direction

$1.90=\dfrac{22.0-\epsilon}{R}$...(II)

From equation (I) and (II)

$\epsilon =2.6\ V$

$R = 12.9\ Omega$

Hence, The value of $\epsilon$ and R are 12.9 ohm and 2.6 V.