To practice Problem-Solving Strategy 25.1 Resistor Circuits. Find the currents through and the potential difference across each resistor in the circuit shown on the diagram (Figure 1) . Use the following values: E = 12.0V , R1 = 15.0Ω , R2 = 45.0Ω , R3 = 20.0Ω , and R4 = 25.0Ω .

Part A

Step by step, reduce the circuit to the smallest possible number of equivalent resistors in order to find the equivalent resistance Req of the entire circuit.

Express you answer in ohms to three significant figures.

Part B

Find Ieq, the current through the equivalent resistor.

Express your answer in amperes to three significant figures.

I₁ = 0.32 A

I₂ = 0.16 A

I₃ = 0.16 A

I₄ = 0.16 A

Explanation:

Part A

The equivalent resistance of the circuit is

Req = R₁ + (R₂||(R₃ + R₄))

Req = 15 + (45||(20 + 25))

Req = 15 + (45||45) = 15 + ((45×45)/(45+45)) = 15 + 22.5 = 37.5 Ω

Part B

From Ohm's law,

V = IR

Ieq = V/(Req) = 12/(37.5) = 0.32 A

Part C

Current through R₁ is the same as Ieq as R₁ is directly in series with the voltage source.

I₁ = 0.32 A

Then, this current flows through the (R₂||(R₃ + R₄)) loop too as the entire loop is in series with R₁

This current is them split into two branches of R₂ and (R₃ + R₄), since these two branches have equal resistances (45 Ω and 45 Ω), 0.32 A is split equally between the R₂ and (R₃ + R₄) branch.

Current through R₂ (using current divider)

I₂ = (45/90) × 0.32 = 0.16 A

Current through (R₃ + R₄) = 0.16 A too.

And because the two resistors are in series, the same current flows through them.

I₃ = I₄ = 0.16 A