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Question

Consider the circuit shown in the figure (32-E19a). Find (a) the current in the circuit (b) the potential drop across the 5 Ω resistor (c) the potential drop across the 10 Ω resistor (d) Answer the parts (a), (b) and (c) with reference to the figure (32-E19b).

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Solution

(a)
Applying KVL in the above loop, we get:

10i+6+5i+12=010i+5i=-1815i=-18i=-1815=-65=-1.2 A

The negative sign indicates that current is flowing in the direction opposite to our assumed direction.
(b) Potential drop across the 5 Ω resistor= 5i = 5×(-1.2 ) = -6 V

(c) Potential drop across the 10 Ω resistor = 10i = (-1.2) × 10 = 12 V

(d)
Applying KVL in the above loop, we get:
10i+5i+6+12=015i=-18i=-1.2 A

Potential drop across the 5Ω register = -6 V
Potential drop across the 10Ω register = -12 V

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