Guided Study: Resistive circuit Working Examples 1. For the circuit shown in Fig.Q1 calculate the values for (a) the current through each resistor, (b) the p.d. across each resistor and (c) the power dissipated by the 20Ω resistor.
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20V

Fig.Q1 Solution:

1 1 1   R AB 25 15

R AB  9.375

1 1 1 1    RBC 10 10 20

RBC  4

R AC  R AB  RBC  13.375

I

20 20   1.495 A R AC 13.375

(a) By using Current Divider Rule

I1 
I2  I3  I4  I5 

R AB 9.375 I   1.495 A  0.56 A 25 25
R AB 9.375 I   1.495 A  0.935 A 15 15 RBC 4  I   1.495 A  0.598 A 10 10 RBC 4  I   1.495 A  0.598 A 10 10 RBC 4 I   1.495 A  0.299 A 20 20

(b) The voltage across reistors 25Ω and 15 Ω is VAB
V AB  I  R AB  1.495  9.375  14.019V

The voltage across reistors 10Ω, 10Ω and 20 Ω is VBC VBC  I  RBC  1.495  4  5.981V (c) The power dissipated by the 20Ω resistor is 2 VBC P20   1.789W 10 or  I 2  20  1.788W or  I  VBC  1.788W

2.

Determine the p.d. between terminals E and F of the circuit in Fig.Q2.

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40V

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E  F

Fig.Q2 Solution: RBF  10  10  25 1 RBCF I  1 1  15 25 RBCF  9.375

R ACF  10  RBCF  19.375 40  2.065 A R ACF

VBCF  I  RBCF  19.355V Or by using Voltage Divider Rule R VBCF  BCF  40  19.355V R ACF Then VEF  10 10  VBCF   19.355  7.742V RBF 25

3.

For the circuit of Fig. Q3 calculate (a) the p.d. across the 8Ω resistor, (b) the current through the 20Ω resistor and (c) the current through the 5Ω resistor.

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30V
Fig.Q3 Solution:
1 1 1   RBC 10 8 RBC  4.444

R AC  5  RBC  9.444 (a) By using Voltage Divider Rule RBC VBC   30  14.116V 5  RBC I1  I2  30  1.5 A 20 30 30   3.177 A R AC 9.444

(b)

(c)

4.

Two resistor of 20Ω and 15Ω are connected in parallel and this combination is connected in series with an unknown resistor Rx. Determine the value of Rx if the complete circuit draws a current of 0.6A from a 12V supply.

Solution:
1 1 1   R AB 20 15 R AC  R AB  R X  RX  R AB  8.571 E I

E  R AB  I 12   8.571  11.429 0.6