Series/Parallel Combination Circuit 2
Solution:
a) Calculate the voltage drop across the parallel branch.
Given that the current is 0.60 A and that the resistance of the branch is 15 W, the voltage drop is
V = (0.60 A)(15 W) = 9.0 V
Thus the voltage drop across the 5.0 W resistor is also 9.0 V, so the current through that branch is
The total current in the circuit is then 2.4 A. (Kirchoff's Current Law)
Now calculate the voltage drops across the 6.0 W and 4.0 W resistors.
V_{6}_{W }= (2.4A)(6.0W) = 14.4V
V_{4}_{W }= (2.4A)(4.0W) = 9.6V
Analysis of the circuit reveals that the terminal voltage of the circuit is equal to the sum of the voltage drops
V_{terminal} = 9.0 V + 14.4 V + 9.6 V = 33 V
b) Emf of the battery is given by
E = V_{terminal} + Ir = 33 V + (2.4A)(1.7W) = 37 V
c) Voltage drop V_{ac} is simply the sum of the voltage drop across the 6.0 W and 7.0 W resistors.
V_{7}_{W }= (0.60A)(7.0W) = 4.2V
V_{ac} = 14.4 V + 4.2 V = 18.6 V
d) Power dissipated by the 5.0 W is given by P = I^{2}R,
P = (1.8A)^{2}(5.0 W) = 16.2 W
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