In the circuit shown, `R_(1) = 4 Omega, R_(2) = R_(3) = 15 Omega, R_(4) = 30 Omega` and E = 10 V. Calculate the equivalent resistance of the circuit and the current in each resistor.

Explanation

The resistance `R_(2) , R_(3) and R_(4)` are connected in parallel in the network. The equivalent resistance between the points A and B will be

`(1)/(R_(AB)) = (1)/(R_(2)) + (1)/(R_(4)) + (1)/(R_(3)) = (1)/(15) + (1)/(30) + (1)/(15)`
`(1)/(R_(AB)) = (4)/(30) + (1)/(30)`
`:. R_(AB) = (30)/(5) = 6Omega`
Total resistance in this network.
`R_(eq) = R_(1) + R_(AB) = 4+6 = 10 Omega`
Current in the circuit will be
`I_(1) = (E)/(R_("eqvt")) = (10)/(10) = 1A`
`:. I_(1) = 1A`
Applying KVL loop in(I) in figure (a), we get
`-I_(1) R_(1) - I_(2)R_(2) + E=0`
`-1 xx 4 -I_(2) R_(2) + 10 = 0`
`I_(2) R_(2) = 6V`
`rArr I_(2) = (6)/(15) = 0.4A`
Voltage across the resistor `R_(2) = 6V` which will be same for resistance `R_(3) and R_(4)` so, `I_(3)R_(3) = 6V`
`rArr I_(3) = (6)/(15) = 0.4A`
Voltage across the resistor `R_(4) = 6V`
`rArr I_(4) = (6)/(30) = (1)/(5) = 0.2A`