Chapter 23, Problem 55A

(a)

To determine

The hottest resistor in the circuit.

Answer to Problem 55A

The hottest resistor in the circuit is $50.0\Omega$ .

Explanation of Solution

Given:

The given circuit is shown below.

The voltage is $V=70\text{ V}$ .

The resistance is $R_1=35.0\text{Ω}$ .

The resistance is $R_2=15.0\text{Ω}$ .

The resistance is $R_3=50.0\text{Ω}$ .

Formula used:

The expression for the power is,

$P=I^{2}R$

Here, $I$ is the current and $R$ is the resistance.

Calculation:

The current $I$ is constant in series circuit. The power of $35.0\Omega$ is,

$\begin{array}{l}P=I^{2}R\\ P_{35.0\Omega }=35I^2\end{array}$

The power of $15.0\Omega$ is,

$\begin{array}{l}P=I^{2}R\\ P_{15.0\Omega }=15I^2\end{array}$

The power of $50.0\Omega$ is,

$\begin{array}{l}P=I^{2}R\\ P_{50.0\Omega }=50I^2\end{array}$

The maximum power will developed by the largest resistor. So, the hottest resistor in the circuit will be the one with larger value of resistance.

Conclusion:

Thus, the hottest resistor in the circuit having the resistance of $50.0\Omega$ .

(b)

To determine

the coolest resistor in the circuit.

Answer to Problem 55A

The coolest resistor in the circuit having the resistance of $15.0\Omega$ .

Explanation of Solution

Given:

The given circuit is shown below.

The voltage is $V=70\text{ V}$ .

The resistance is $R_1=35.0\text{Ω}$ .

The resistance is $R_2=15.0\text{Ω}$ .

The resistance is $R_3=50.0\text{Ω}$ .

Formula used:

The expression for the power is,

$P=I^{2}R$

Here, $I$ is the current and $R$ is the resistance.

Calculation:

From part (a),

The power of $35.0\Omega$ is $P_{35.0\Omega }=35I^2$ .

The power of $15.0\Omega$ is $P_{15.0\Omega }=15I^2$ .

The power of $50.0\Omega$ is $P_{50.0\Omega }=50I^2$ .

The least power will developed by the smallest resistor.

Conclusion:

Thus, the coolest resistor in the circuit having the resistance of $15.0\Omega$ .

(c)

To determine

The current in ammeter.

Answer to Problem 55A

The current in ammeter is $2.0\text{ A}$ .

Explanation of Solution

Given:

The given circuit is shown below.

The voltage is $V=70\text{ V}$ .

The resistance is $R_1=35.0\text{Ω}$ .

The resistance is $R_2=15.0\text{Ω}$ .

The resistance is $R_3=50.0\text{Ω}$ .

Formula used:

The expression for the current by Ohm’s law is,

$I=\frac{V}{R}$

Here, $V$ is the voltage and $R$ is resistance.

Calculation:

The current in ammeter is,

$\begin{array}{l}I=\frac{V}{R_1}\\ I=\frac{\text{70 V}}{\text{35}\text{.0 }\Omega }\\ I=2.0\text{ A}\end{array}$

Conclusion:

Thus, the current in ammeter is $2.0\text{ A}$ .

(d)

To determine

The power supplied by the battery.

Answer to Problem 55A

The power supplied by the battery is $4\times {10}^2\text{ W}$ .

Explanation of Solution

Given:

The given circuit is shown below.

The voltage is $V=70\text{ V}$ .

The resistance is $R_1=35.0\text{Ω}$ .

The resistance is $R_2=15.0\text{Ω}$ .

The resistance is $R_3=50.0\text{Ω}$ .

Formula used:

The expression for the power is,

$P=I^{2}R$

Here, $I$ is the current and $R$ is the resistance.

Calculation:

From part (c), the current in ammeter is $I=2.0\text{ A}$ .

The total resistance is,

$\begin{array}{l}R=R_1+R_2+R_3\\ R=35.0\Omega +15.0\Omega +50.0\Omega \\ R=1\times {10}^2\Omega \end{array}$

The power supplied by the battery is,

$\begin{array}{l}P=I^{2}R\\ P={\left(2.0\text{ A}\right)}^2\left(\text{1}\times {\text{10}}^2\Omega \right)\\ P=4\times {10}^2\text{ W}\end{array}$

Conclusion:

Thus, the power supplied by the battery is $4\times {10}^2\text{ W}$ .