In the given circuit

(Choose the incorrect option).

Options

A. The current through the `10 Omega` resistor is `2 A`
B. The current through the `5 Omega` resistor is `2.5 A`
C. The current through the `25 V` battery is `6.25 A`
D. the `p.d` between the terminals of `10 V` battery is `12.5 V`

Answer

A

Explanation

`E = ((10)/(1)+(25)/(2))/((1)/(1)+(1)/(2)) = (45)/(3) = 15 V`
`r = (1 xx 2)/(1 + 2) = (2)/(3) Omega`

`i = (15)/((2)/(3)+(10)/(3)) = (15)/(4) A`
`V_A - V_B = i xx (10)/(3) = (15)/(4) xx (10)/(3) = 12.5 V`
Current in `10 Omega, (V_A - V_B)/(10) = (12.5)/(10) = 1.25 A`
Current in `5 Omega`, `(V_A - V_B)/(5) = (12.5)/(5) = 2.5 A`
Current in `25 V` battery, `V_A - V_B = 25 - 2 i_0`
`12.5 = 25 - 2 i_0`
`i_0 = 6.25 A`.