Analyze the circuit in the figure ( R1 = R2 = R3 = 1080 ). R1 is connected in se

Question

Analyze the circuit in the figure ( R1 = R2 = R3 = 1080 ). R1 is connected in series to a parallel system of R2 and R3. a) Express the resistors in parallel (R3 and R3) as a single equivalent resistor. What is the value of this single resistor? b) Combine this equivalent resistor with R1 to get the total equivalent resistance of the circuit. What is the value of this resistor? c) The circuit now is reduced to a battery in series with a single equivalent resistance. Use your result from the previous question to find the current through the battery ( E = 10.2 V) and resistor R1.

Explanation / Answer

Part A)

Resistors in parallel add as 1/Req = 1/R1 + 1/R2

1/Req = 1/1080 + 1/1080

Req = 540

Part B)

Resistors in series add as Req = R1 + R2

Req = 1080 + 540 = 1620

Part C)

Apply V = IR

10.2 = (I)(1620)

I = 6.30 X 10-3 A which is 6.30 mA

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