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example: 3.1-3
The power dissipated in the resistor marked with * in the circuit in 
Fig. 3.1-6 is found to be 12.5 W. (i) Find the value of R (ii) Find the 
total power delivered by the source. (iii) Find the voltage, current 
and power dissipated for all the resistors.
Solution
(i) We need to find an expression for power dissipated in the 
marked resistor in terms of R. We need an expression for 
current in it for that. We use series-parallel reductions 
successively as shown in Fig. 3.1-7.
Therefore, the current delivered by the voltage source is 20/2R 
=
10/R. The series-parallel 
reduction stages show that this current divides into two equal components at the node A since 
equivalent resistance to the right of A is 2R. Therefore, 5/R A reaches node B and splits into 2.5/R 
A into the two equal resistors connected in parallel there. Therefore, the current in the marked 
resistor is 2.5/R. Therefore, power dissipated in it is 2R 
× 
6.25/R
2
=
12.5/R. Since this is stated to 
be 12.5 W, the value of R 
=
1 
W
.
20 V
R
R
R
2
R
+
–
20 V
R
2
R
2
R
+
–
20 V
R
R
+
–
Fig. 3.1-7 
Series – parallel reduction for Example 3.1-3 
(ii) The equivalent resistance across the 20V source is 2R 
=
2 
W
. Therefore, the current delivered by 
the source is 10 A and the power delivered by the source is 20 
× 
10 
=
200 W.
20 V
R
R
2
R
2
R
2
R
A
B
+
–
Fig. 3.1-6 
Circuit for 
Example 3.1-3
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The Resistor 
3.7
(iii) The various currents, voltages and power values are marked in Fig. 3.1-8. 
20 V
10 V
5 V
5 V
5 V
5 A
5 A
10 A 10 V
1 
Ω
1 
Ω
2 
Ω
2 
Ω
2 
Ω
2.5 A
2.5 A
(50 W)
(12.5 W) (12.5 W)
(100 W)
(25 W)
+
+
+
+
+
+
−
−
−
−
−
−
Fig. 3.1-8 
Circuit solution for Example 3.1-3