A series-parallel circuit consists of two parallel circuits connected in series across a 45-V source. One parallel branch consists of an R1 of 17 kΩ and an R2 of 23 kΩ. The other parallel branch consists of an R3 of 45 kΩ and an R4 of 55 kΩ. The current (LaTeX: mAm A) through R4 is:

The answer to the question is

The current  through R4 = 0.5865 mA

Explanation:

To solve this we list out the known thus

Voltage source = 45-V

Firsrt parallel circiuit has resistances of R1  = 17 kΩ and R2 = 23 kΩ

Second parallel circiuit has resistances of R3  = 45 kΩ and R4 = 55 kΩ

we first find the current flowing in the circuit by finding thr sum of the total resistance in eah parallel circuit

Firsrt parallel circiuit, for circuit in parallel, sum of resistance 1/RT1 = 1/ R1 + 1/R2 = 1/17+1/23 =  0.1023017 therefore RT1 = 1/0.1023017 = 9.775 kΩ

Similarly we have for the second parallel circuit 1/RT2 = 1/R3 + 1/R4

= 1/45 + 1/55 =  0.0404 Hence RT2 = 1/0.0404 = 24.75 kΩ

This means that the 9.775 kΩ and the 24.75 kΩ are in series hence total resistance of the circuit = sum of all resistances in series  

= 9.775 kΩ + 24.75 kΩ = 34.525 kΩ

However current, I is given by V/R = 45-V/34.525 kΩ = 1.303 × 10⁻³ A or 1.303 mA

From the current divider rule, I4 = I × (R4/(R3+R4)

That is the currrent flowing throuhgh R4 = 1.303  × (45/(45+55)) = 0.5865 mA

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