Question

In the circuit shown in fig. the cell has emf 10V and internal resistance `1 Omega`

A. The current through the `3Omega` resistor is 1A.
B. The currents through the `3 Omega` resistor is 0.5 A.
C. The currents through the `4 Omega` resistor is 0.5 A
D. The current through the `4Omega` resistor is 0.25 A.

Answer 1

Correct Answer - A::D
The last three resistors `2 Omega, 4 Omega` and `2 Omega` are in series having
the equivalent resistance as `2+4+2 = 8 Omega`. This will be in
series with the `8 Omega` next to them. So their equivalent resistance
becomes `4 Omega` in this way, net equivalent resistance of the
circuit becomes `R_(eq) =9 Omega` This will be in series with `r = 1Omega`.
So the current through `3 Omega` is.
`I = e(R +r) = 10//(9+1) =1A`
Further, current will get divided at C and E into half at each
point. So finally the current reaching in `4 Omega` will be 0.25A.