Chapter 2, Problem 54E

For each of the circuits in Fig. 2.40, find the current I and compute the power absorbed by the resistor.

FIGURE 2.40

To determine

Find the current and power absorbed by resistor for each circuit in the given Figures.

Answer to Problem 54E

The current flowing in the circuit in Figure 1 is $500\text{ μA}$,

The power absorbed by the resistor in Figure 1 is $2.5\text{ mW}$,

The current flowing in the circuit in Figure 2 is $-500\text{ μA}$,

The power absorbed by the resistor in Figure 2 is $2.5\text{ mW}$,

The current flowing in the circuit in Figure 3 is $-500\text{ μA}$,

The power absorbed by the resistor in Figure 3 is $2.5\text{ mW}$,

The current flowing in the circuit in Figure 4 is $500\text{ μA}$ and

The power absorbed by the resistor in Figure 4 is $2.5\text{ mW}$

Explanation of Solution

Formula used:

The expression for current is as follows.

$I=\frac{v}{R}$ (1)

Here,

$I$ is current flowing in the circuit,

$v$ is Voltage given in the circuit and

$R$ is Value of resistance in the circuit.

The expression for power absorbed by the resistor is as follows.

$p=v\times I$ (2)

Here,

$p$ is power absorbed by the resistor,

Calculation:

Refer to FIGURE 2.40(a) in the textbook.

The circuit diagram is redrawn as shown in Figure 1.

Substitute $5\text{ V}$ for $v$ and $10\text{ kΩ}$ for $R$ in equation (1).

$\begin{array}{c}I=\frac{5\text{ V}}{10\text{ kΩ}}\\ =\frac{5\text{ V}}{10\times 1000\text{ Ω}}\left\{\because 1\text{ kΩ}=\text{1000 Ω}\right\}\\ =500\times {10}^{-6}\text{ A}\\ =\text{500 μA }\left\{\because {10}^{-6}\text{ A}=\text{1 μA}\right\}\end{array}$

Substitute $5\text{ V}$ for $v$ and $500\text{ μA}$ for $I$ in equation (2).

$\begin{array}{c}p=\left(5\text{ V}\right)\times \left(500\text{ μA}\right)\\ =\left(5\text{ V}\right)\times \left(\text{500}\times {\text{10}}^{-6}\text{ A}\right)\left\{\because 1\text{ μA}={\text{10}}^{-6}\text{ A}\right\}\\ =2.5\times {10}^{-3}\text{ W}\\ =\text{2}\text{.5 mW }\left\{\because {10}^{-3}\text{ W}=\text{1 mW}\right\}\end{array}$

Refer to FIGURE 2.40(b) in the textbook.

The circuit diagram is redrawn as shown in Figure 2.

Refer to redrawn Figure 2

Substitute $-5\text{ V}$ for $v$ and $10\text{ kΩ}$ for $R$ in equation (1).

$\begin{array}{c}I=\frac{-5\text{ V}}{10\text{ kΩ}}\\ =\frac{-5\text{ V}}{10\times 1000\text{ Ω}}\left\{\because 1\text{ kΩ}=\text{1000 Ω}\right\}\\ =-500\times {10}^{-6}\text{ A}\\ =-\text{500 μA }\left\{\because 1\text{ μA}={\text{10}}^{-6}\text{ A}\right\}\end{array}$

Substitute $-5\text{ V}$ for $v$ and $-500\text{ μA}$ for $I$. in equation (2).

$\begin{array}{c}p=\left(-5\text{ V}\right)\times \left(-500\text{ μA}\right)\\ =\left(-5\text{ V}\right)\times \left(-\text{500}\times {\text{10}}^{-6}\text{ A}\right)\left\{\because 1\text{ μA}={\text{10}}^{-6}\text{ A}\right\}\\ =2.5\times {10}^{-3}\text{ W}\\ =\text{2}\text{.5 mW }\left\{\because {10}^{-3}\text{ W}=\text{1 mW}\right\}\end{array}$

Refer to FIGURE 2.40(c) in the textbook.

The circuit diagram is redrawn as shown in Figure 3.

Refer to redrawn Figure 3.

Substitute $-5\text{ V}$ for $v$ and $10\text{ kΩ}$ for $R$ in equation (1).

$\begin{array}{c}I=\frac{-5\text{ V}}{10\text{ kΩ}}\\ =\frac{-5\text{ V}}{10\times 1000\Omega }\left\{\because 1\text{ kΩ}=\text{1000 Ω}\right\}\\ =-500\times {10}^{-6}\text{ A}\\ =-\text{500 μA }\left\{\because 1\text{ μA}={\text{10}}^{-6}\text{ A}\right\}\end{array}$

Substitute $-5\text{ V}$ for $v$ and $-500\text{ μA}$ for $I$. in equation (2).

$\begin{array}{c}p=\left(-5\text{ V}\right)\times \left(-500\text{ μA}\right)\\ =\left(-5\text{ V}\right)\times \left(-\text{500}\times {\text{10}}^{-6}\text{ A}\right)\left\{\because 1\text{ μA}={\text{10}}^{-6}\text{ A}\right\}\\ =2.5\times {10}^{-3}\text{ W}\\ =\text{2}\text{.5 mW }\left\{\because {10}^{-3}\text{ W}=\text{1 mW}\right\}\end{array}$

Refer to FIGURE 2.40(d) in the textbook.

The circuit diagram is redrawn as shown in Figure 4.

Refer to redrawn Figure 4.

Substitute $5\text{ V}$ for $v$ and $10\text{ kΩ}$ for $R$ in equation (1).

$\begin{array}{c}I=\frac{5\text{ V}}{10\text{ kΩ}}\\ =\frac{5\text{ V}}{10\times 1000\text{ Ω}}\left\{\because 1\text{ kΩ}=\text{1000 Ω}\right\}\\ =500\times {10}^{-6}\text{ A}\\ =\text{500 μA }\left\{\because {10}^{-6}\text{ A}=\text{1 μA}\right\}\end{array}$

Substitute $5\text{ V}$ for v and $500\text{ μA}$ for $I$ in equation (2).

$\begin{array}{c}p=\left(5\text{ V}\right)\times \left(500\text{ μA}\right)\\ =\left(5\text{ V}\right)\times \left(\text{500}\times {\text{10}}^{-6}\text{ A}\right)\left\{\because 1\text{ μA}={\text{10}}^{-6}\text{ A}\right\}\\ =2.5\times {10}^{-3}\text{ W}\\ =\text{2}\text{.5 mW }\left\{\because {10}^{-3}\text{ W}=\text{1 mW}\right\}\end{array}$

Conclusion:

Thus, The current flowing in the circuit in Figure 1 is $500\text{ μA}$,

The power absorbed by the resistor in Figure 1 is $2.5\text{ mW}$,

The current flowing in the circuit in Figure 2 is $-500\text{ μA}$,

The power absorbed by the resistor in Figure 2 is $2.5\text{ mW}$,

The current flowing in the circuit in Figure 3 is $-500\text{ μA}$,

The power absorbed by the resistor in Figure 3 is $2.5\text{ mW}$,

The current flowing in the circuit in Figure 4 is $500\text{ μA}$ and

The power absorbed by the resistor in Figure 4 is $2.5\text{ mW}$