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

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

Step-by-step 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=(V)/(R)`

`1.50=(\epsilon+22.0)/(R)`...(I)

In reverse direction

`1.90=(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.