Question

In Fig. 27-25, the ideal batteries have emfs ${\mathbf{\epsilon }}_{\mathbf{1}}\mathbf{=}\mathbf{12}\mathbf{v}$and ${\mathbf{\epsilon }}_{\mathbf{2}}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{0}\mathbf{v}$. What are (a) the current, the dissipation rate in (b) resistor 1?$\mathbf{\left(}\mathbf{4}\mathbf{\Omega }\mathbf{\right)}$And (c) resistor 2 $\mathbf{\left(}\mathbf{8}\mathbf{\Omega }\mathbf{\right)}$, and the energy transfer rate in (d) battery 1 and (e) battery 2? Is energy being supplied or absorbed by (f) battery 1 and (g) battery 2?

Short answer

1. Current in the circuit is 0.50A.
2. Dissipation rate in resistor 1 $\left(4\Omega \right)$is 1W.
3. Dissipation rate in resistor 2$\left(8\Omega \right)$is 2W.
4. Energy transfer rate in battery 1 is 6W.
5. Energy transfer rate in battery 2 is 3W.
6. Battery 1 supplies the energy to the circuit.
7. Battery 2 absorbs the energy from the circuit.

Step-by-step solution

Step 1: Given

Emf,${\epsilon }_1=12V$

Emf,${\epsilon }_2=6V$

Resistance,$R_1=4\Omega$

Resistance,$R_2=8\Omega$

Determining the concept

Using Kirchhoff’s loop law, find the current flowing through the circuit. Using the formula for power, find the rate of dissipation through the given resistors. Using the equation of the rate of energy transfer, find the energy transfer rate of the battery.

Kirchhoff's loop rule states that the sum of all the electric potential differences around a loop is zero.

Formulae are as follow:

$$\begin{gathered} P=i^{2}R \\ P=i\epsilon \end{gathered}$$

Where, P is power, I is current, R is resistance, role="math" localid="1662570355324" $\epsilon$is emf.

(a) determining the current in the circuit

The current in the circuit

Let be the current in the circuit and take it to be positive if it is to the left in R₁.

Applying the Kirchhoff’s loop law,

${\epsilon }_1-i{R}_2-i{R}_1-{\epsilon }_2=0$

Solving for the current, gives,

$i{R}_1+i{R}_2={\epsilon }_1-{\epsilon }_2$

Taking i common in the left side,

$i(R_1+R_2)={\epsilon }_1-{\epsilon }_2$

$i=\frac{({\epsilon }_1-{\epsilon }_2)}{(R_1+R_2)}$

$i=\frac{(12V-6V)}{(4\Omega +8\Omega )}$

i=0.50A

Hence, Current in the circuit is 0.50A.

(b) determining the dissipation rate in resistor 1(4Ω)

The dissipation rate in resistor 1,$\left(4\Omega \right)$

The dissipation rate in the resistor is given by the formula,

$P=i^{2}R$

For resistor,

$$\begin{gathered} P_1=i^2{R}_1 \\ {P}_1=(0.50A{)}^2(4\Omega ) \\ {P}_1=1.0W \end{gathered}$$

Hence, Dissipation rate in resistor 1 $\left(4\Omega \right)$is 1W.

(c) Determining the dissipation rate in resistor 2(8Ω).

The dissipation rate in resistor 2$\left(8\Omega \right)$:

$$\begin{gathered} P_2=i^2{R}_2 \\ {P}_2=(0.50A{)}^2(8\Omega ) \\ {P}_2=2.0W \end{gathered}$$

Hence, dissipation rate in resistor 2 $\left(8\Omega \right)$is 2W.

(d) determining the energy transfer rate in battery 1

The energy transfer rate in battery 1

For battery 1 emf is${\epsilon }_1$and the rate of energy transfer is given by,

$P_1=i{\epsilon }_1=(0.50A)\left(12V\right)=6.0W$

Hence, energy transfer rate in battery 1 is 6W.

(e) determining the energy transfer rate in battery 2

The energy transfer rate in battery 2:

For battery 2emf is ${\epsilon }_2$and the rate of energy transfer is given by,

$P_2=i{\epsilon }_2=(0.50A)\left(6V\right)=3.0W$

Hence, energy transfer rate in battery 2 is 3W.

(f) Determining is energy being supplied or absorbed by battery 1

The energy is supplied only when the emf and the current flowing inside the battery are in the same direction.

In battery 1 the current is in the same direction as the emf.

Therefore, this battery supplies the energy to the circuit; the battery is discharging.

(g) Determining is energy being supplied or absorbed by battery 2

The battery absorbs the energy at the rate $P=i\epsilon$only if the current and the emf are in the opposite directions.

The current in battery 2 is in opposite the direction of the emf, so, this battery absorbs the energy from the circuit. It is charging.

Hence, the battery 2 absorbs the energy from the circuit.

Therefore, using the Kirchhoff’s loop law, find the current flowing through the circuit and using the equation of power, find the dissipation rate of resistor and the energy transfer rate of the battery.