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Question
Consider the circuit shown in the figure. 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.
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Solution
(a)
Applying KVL in the above loop, we get:-
\[10i + 6 + 5i + 12 = 0\]
\[ \Rightarrow 10i + 5i = - 18\]
\[ \Rightarrow 15i = - 18\]
\[ \Rightarrow i = - \frac{18}{15} = - \frac{6}{5} = - 1 . 2 A\]
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:-
\[10i + 5i + 6 + 12 = 0\]
\[ \Rightarrow 15i = - 18\]
\[ \Rightarrow i = - 1 . 2 A\]
Potential drop across the 5Ω register = -6 V
Potential drop across the 10Ω register = -12 V
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