A 20 V battery of internal resistance `1 Omega` is connected to three coils of `12Omega, 6 Omega, and 4Omega` in parallel, a resistor of `5 Omega` and a reversed battery (emf 8 V and internal resistance 2 Omega) as shown in fig Calculate

a. the current in the circuit,
b. current in resistor of `12 Omega` coil, and
c. potential difference across each battery.

Explanation

Equivalent resistance R' of the parallel combination of `12 Omega, 6 Omega, and 4 Omega` resistance is given by
`(1)/(R') = (1)/(12) +(1)/(12) +(1)/(6) +(1)/(4) or R' = 2Omega`
Total nresistance of the circuit is `R = 1+5 +2 + 2 = 10 Omega`
E = 20 - 8 = 12 V
a. Current in the circuit is ` I = (E )/(R ) = (12)/(10) = 1.2A`
b. Combined resistance of `6 Omega and 4 Omega` in parallel is
`(6xx4)/(6+4) = 2.4 Omega`
current in resistor of `12 Omega` coil is `(2.4)/(12 +2.4)xx1.2 = 0.2A`
c. Potential difference across the 20 V battery is `(20 - 1.2xx1)`
=18.8 V
Potential difference across the 8 V battery is `(8 +1.2 xx2)`
=10.4 V