Find out the folowing in the electric circuit given in (Fig. 3.48).
(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.
.

Explanation

(a) Since, two `8Omega` resistors are in parallel, then their effective resistance `R_(P)` is given by
`(1)/(R_(p)) = (1)/(R_(1)) + (1)/(R_(2))`
`(1)/(R_(p)) = (1)/(8) + (1)/(8) = (1)/(4) Omega`
(b) Total resistance in the circuit `R = 4 Omega + R_(p)`
`= 4 Omega + 4 Omega`
`= 8 Omega`
Current through the circuit `I=(V)/(R) = (8)/(8) = 1A`
`" "{:(("where,", V = " Voltage of battery and"),(,R = "Net equivalent resistance of circuit")):}`
Thus, current through `4 Omega` resistor is 1 A as `4 Omega` and `R_(p)` are in series and same current flows through them.
(c) Potential difference across `4 Omega` resistor is potential drop by the `4 Omega` resistor i.e, `" "V = IR = 1 xx 4 = 4 V`
(d) Power dissipated in `4 Omega` resistor `P = I^(2)R = 1^(2) xx 4 = 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 current.