In the circuit shown , `R_1=4Omega, R_2=R_3 =15Omega, R_4 = 30 Omega ` and `epsi = 10V` . Calculate the equivalent resistance of the circuit and the current in each resistor.

Explanation

The given circuit can be redrawn as shown in Fig. Here `R_2 , R_3` and `R_4` are in parallel and their combined resistance R. will be
`(1)/(R.) = (1)/(R_2) + (1)/(R_3) + (1)/(R_4) = 1/15 + 1/15 + 1/30 = (2 + 2 + 1)/(30) = 5/30`
` rArr R. = 6Omega`
Now R. and `R_1` are in series, hence equivalent resistance of the circuit
`R = R. + R_1 = 6Omega +4Omega = 10 Omega`
Main circuit current `I_1 = E/R = (10V)/(10 Omega) = 1 A `
`I_2 + I_3 + I_4 = I_1 = 1A`
and potential difference between A and B remains constant, i.e.,
`15I_2 = 15I_3 = 30 I_4`
` rArr I_2 = I_3 = 2I_4`
Therefore, on simplifying, we get
`I_2 = I_3 = 0.4A` and `I_4 = 0.2 A `