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 figure. 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 `12 Omega, 6 Omega` and `4 Omega` connected in parallel is
`1/R' =1/12 + 1/6 + 1/4 = 6/12=1/2 or R' = 2 Omega`
Total resistance of circuit, `R = 1 +5 +2+2`
`= 10 Omega`
Effective emf of circuit `= 20 -8 = 12V`
Current in the circuit `I=12/10 =1.2 A`
So, current through each battery and `5 Omega` resistor, `= 1.2 A`
Pot. diff. across the parallel combination of three resistors, `V' = IR' = 1.2 xx2 =2.4V`
Current in `12 Omega` coil = `2.4/12=0.2A`
Current in `6 Omega` coil = `2.4/6=0.4A`
Current in `4 Omega` coil = `2.4/4=0.6A`
Terminal pot. diff. across 20 V battery
`V = epsilon - Ir = 20 -1.2 xx1 = 18.8 V`
Terminal pot. diff. across8 V battery
`V'= epsilon' + Ir' = 8 +1.2 xx 2=10.4V`