Consider the following figure. (Assume R 1 = 29.0 ?, R 2 = 18.0 ?, and V = 18.0

Question

Consider the following figure. (Assume R1 = 29.0 ?, R2 = 18.0 ?, and V = 18.0 V.)

(a) Can the circuit shown above be reduced to a single resistor connected to the batteries? Explain.

(b) Find the magnitude of the current and its direction in each resistor (A).

18.0 ohms

5.0 ohms

29.0 ohms

Consider the following figure. (Assume R1 = 29.0 ?, R2 = 18.0 ?, and V = 18.0 V.) (a) Can the circuit shown above be reduced to a single resistor connected to the batteries? Explain. (b) Find the magnitude of the current and its direction in each resistor (A). 18.0 ohms 5.0 ohms 29.0 ohms

Explanation / Answer

a)

No ,Since Multiple loop circuits does not have any resistors in series nor in parallel.Thus the circuit cannot be simplified further.

b)

Let I1=Current flowing through R1

I2=Current flowing through 5 ohms

I3=Current flowing through R2

From Kirchoff's Juction rule

I3=I1+I2

I1+I2-I3=0--------------------------1

Applying Kirchoff's loop rule on the upper loop ,so

V-29I1+5I2-10=0

29I1-5I2=8--------------------------2

Applying kirchoff's loop rule on the lower loop ,so

10-5I2-18I3=0

5I2+18I3=10------------------------3

Solving 1,2 and 3 we get

I3=I18=0.502 A

I2=I5=0.1929 A

I1=I29=0.3091 A

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