I2 A) In the circuit diagram shown currents I, I2, and I are unknown 9.0 ? 4012

Question

I2 A) In the circuit diagram shown currents I, I2, and I are unknown 9.0 ? 4012 Complete the table below as you write Kirchhoffs rules for the nodes and loops specified. Use the conventional direction of current. (10 points) 13 V 24 V Node d (2 points) Loop adcba (countercklockwise) (3 points) Loop defed (clockwise) (3 points) Suppose the solution gave you: h = 2 A and 12--??. What do these values mean for the circuit above? (2 points) B) For the circuit shown in the figure below. I-1.3 A and R-15 ?, what is the value of the emf ? Use combination of resistors for the solution, not Kirchhoft's loop rule.(10 points) 2R 2R

Explanation / Answer

1. (a) At node d, current I1 and I2 will enter while I3 will leave the node d. So required equation is,

I1 + I2 = I3

(b) Applying Kirchoff's voltage rule to the loop adcba, we have,

I1×5 - I2×9 = -24 => I2×9 - I1×5 = 24

(c) Applying Kirchoff's voltage rule to the loop dcfed, we have,

I1×5 + I3×4 = 13

(d) If the solution come out to be I1 = 2A and I2 = -3A, it means that the direction of current I1 is actually same as shown in fig. But as the value of I2 = -3A, it means that the actual current direction in the circuit should be opposite to the direction of I2 as in fig.

2. Here the two resistors 2R and 2R are in parallel, their equivalent resistance is R' = 2R×2R/2R + 2R = R

Now R' and R are in series so the equivalent resistance of the circuit is,

Req = R + R' = R+R = 2R = 2×15 = 30 ohm

Hence, Emf, E = IReq = (1.3×30) volt = 39 volt.

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