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).

Answer 1

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

$$\begin{gathered} 10i+6+5i+12=0 \\ \Rightarrow 10i+5i=-18 \\ \Rightarrow 15i=-18 \\ \Rightarrow i=-\frac{18}{15}=-\frac{6}{5}=-1.2\mathrm{A} \end{gathered}$$

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:
$$\begin{gathered} 10i+5i+6+12=0 \\ \Rightarrow 15i=-18 \\ \Rightarrow i=-1.2\mathrm{A} \end{gathered}$$

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