FB 02. Each cell of the circuit shown to the right has an emf of 1.50 V. Problem

Question

FB 02. Each cell of the circuit shown to the right has an emf of 1.50 V. Problem 02 a.) What is the net emf of the circuit? b.) What is the net resistance of the circuit? c.) Determine the current through each resistor. d.) What is the voltage, V2, across resistor R2? e.) What is the power consumption of resistor Ry in this circuit? 03. Assume all of the light bulbs in the circuit shown below are identical, each with a resistance of R a.) What is the net resistance of the circuit? b.) Determine the current through each light bulb. c.) What is the voltage, Vs, across light bulb L4? d.) What is the power consumption of light bulb Ls in this circuit? Problem 03 1 2 3 L5 ANSWERS: 01 . a.) 6.00 V b.) 27.0 c.) 0.222 A 01. d.) 1.-Is-0.222 A, 1-0.0409 A, l,-0.1 11 A, 14-00707 A e.) 156 V ) 2.00 w g) 0667 w 01.h.) 1.33 w i.) 0.172 W j.) 0.1 10w k.) 1.56 V L)4.22 V 02.a.) 12.0 V b.)19.5 02. c.) 0.615 A,-0.461 A, l-4-0.154 A d.) 8.30 V e.) 0.520 w 03. a.)

Explanation / Answer

(a) Va = 1.5 Volts X 6 = 9V

Vb = 1.5 v X 2 = 3V

Net emf in the circuit = 12V

(b) R3 + R4 = 32 +22 =54 ohms

This 54 is parallel with 18 ohms

So equivalent resistance = 54 X 18/(54+18) = 27/2 Ohms

This 27/2 is series with 6 ohms =( 27/2 ) + 6 = 19.5 ohms

(c) Currrent through equivalent resistor 27/2 ohms and 6 ohms is same and is given by

I = V/R = 12 V / 19.5 ohms = 0.615 Amps

So I1 = 0.615 amps

Now 0.615 branches into two current one strem flowing through R3 and R4 and other flowing through R2

but voltage across these streams is same.

So I ( 54 ohms ) = I' X 18 ohms , where I is current through R3 and R4 and I' is current through R2

I/I' = 18/54 = 1/3

we know I +I' = 0.615

I + 3xI = 0.615

I = 0.615/4 = 0.154 Amps

I' = 3 X I = 0.462 Amps

(d) voltage across resistor R2 = 0.462 amps X 18 ohms =8.30 Volts

(e) Power in R3 = I^2 X R3 =0.154^2 X22 ohms = 0.520 watts

Note: I am allowed to answer only one question with four sub-parts but I am answering 5 sub-parts as special case

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