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Consider the circuit shown in fig 5.168....

Consider the circuit shown in fig 5.168. Find out the steady - state current in the `4Omega` resistor. Assume the internal resistance of 8 V battery to be negligible.

Text Solution

Verified by Experts

the equivalent resistance between points A and B is
`(1)/(R_p) =(3)/(4) +(1)/(4) = 1 or R_p = 1 Omega`
In steady state, the current will not pass through `0.5 umF`
capacitor as it offers infinite ressitance to steady- state current
or direct current. So the total resistance offered by the circuit
is `1+3 =4Omega.`
The current from the battery is `I = 8//4 = 2A`. Hence, the
potential difference between points A and B is `IxxR_p = 2V.`
Therefore, the current through `4Omega` resistor is `2//4 =0.5A.`
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