Two cells, `E_(1)` and `E_(2)` of emfs 4 V and 8 V having internal resistances `0.5 Omega` and `1.0 Omega` respectively are connected in opposition to each other. This combination is connected in opposition to with resistance of `4.5 Omega` and `3.0 Omega`. Another resistance of `6 Omega` is connected in parallel across the `3 Omega` resistor. (a) Draw the circuit diagram (b) Calculate the total current flowing through the circuit. (c ) Terminal potential difference acrss cell `E_(1)` and `E_(2)`.

Explanation

(a) The circuit diagram is shown in figure
(b) As the resistance of `3 Omega` and `6 Omega` are in parallel, their effective resistance is
`R' = (3 xx 6)/(3+6) = 2 Omega`
Total resistance of the circuit
`= 4.5 + 2 +1.0 +0.5 = 8 Omega`
Effective emf `= 8 - 4= 4V`
Current `I=4/8 = 0.5A`
(c ) Here, cell `E_(1) is being charged by cell `E_(2)`. So terminal potential difference across cell `E_(1)`
`epsilon_(1) +I r_(1) = 4 +0.5 xx 0.5 = 4.25 V`.
Here cell `E_(2)` is supplying the current in the circuit.
So terminal potential difference across cell `E_(2)`
`=epsilon_(2) - I r_(2) = 8 -0.5 xx 1 = 7.5 V`