Question

Find out the folowing in the electric circuit given in (Fig.
(a) Effective resistance of two `8 Omega` resistors in the combination
(b) Current flowing through `4 Omega` resistor
( c) Potential difference across `4 Omega` resistance
(d) Power dissipated in `4 Omega` resistor
(e) Difference in ammeter readings, if any.
.

Answer 1

(a) Since two `8 Omega` resistors are in parallel, their effective resistance `(R_p)` is given by
`(1)/(R_p)=(1)/(8)+(1)/(8)=(1)/(4)` or `R_p = 4 Omega`
(b) Total resistance in the circuit, `R = 4 Omega + R_p = 4 Omega + 4 Omega = 8 Omega`
Current through the electric circuit, `I = (V)/( R) = (8 V)/(8 Omega) = 1 A`
Since `4 Omega` resistor and `R_p` are in series, current through `4 Omega` resistors `= 1 A`
(c) pd across `4 Omega` resistors, `V = IR = 1 xx 4 = 4 V`
(d) Power dissipated in `4 Omega` resistors, `P = I^2 R = (1 A)^2 (4 Omega) = 4 W`
(e) There is no difference in the readings of ammeters `A_1` and `A_2` as same current flows through all elements in a series curcuit.