A battery of emf `12V` and internal resistance `2 Omega` is connected two a `4 Omega` resistor. Show that the a voltmeter when placed across cell and across the resistor in turn given the same reading

Answer

`8V`

Explanation

Here resistance of circuit `= r+ R = 2+ 4 = 6 Omega`
Current in circuit `I = (epsilon)/( r + R) = (12)/(6) = 2A`
When voltmeter is parallel across the cell then terminal pot diff `V = epsilon - lr = 12 - 2 xx 2 = 8V`
When voltmeter is parallel across the resistor `R` Then pot diff across `R , V' = IR = 2 xx 4= 8V`
As `V = = V'`, hence proved