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If in figure, R1 = 10 Omega,R2 =40 Omeg...

If in figure, `R_1 = 10 Omega,R_2 =40 Omega, R_3 = 30 Omega, R_4 = 20 Omega, R_5 = 60 Omega` a 12 V battery is connected to the arrangement. Calculate ( a ) the total resistance in the circuit and ( b ) the total current flowing in the circuit.

Text Solution

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( a ) Suppose, we replace the parallel resistors `R_1 and R_2` by an eqUivalent resistor of resistance, R.. Similarly we replace the parallel resistors `R_3,R_4 and R_5` by an eqUivalent single resistor of resistance R". Then using equation of equivalent resistance of parallel combination of resistances,
we have `1/(R.) = 1/10 +1/40 =5/40 , " that is " R. = 8 Omega`

Similarly, `1/(R.) = 1/30 + 1/20 + 1/60 + 6/60` , that is,`R.. =1 0Omega`
Thus, the total resistance,R = R. + R.. = `18 Omega`
( b ) To calculate teh current, we use the Ohm.s law, and get
`I = V/R =(12V)/(18Omega) = 0.67 A`.
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