Question

The batteries shown in the circuit in the figure have negligibly small internal resistances.

a) Find the current through the 30.0 ${\rm \Omega}$ resistor.

b) Find the current through the 20.0 ${\rm \Omega}$ resistor.

c) Find the current through the 10.0 ${\rm V}$ battery.

I found the answers for part a and b to be 0.333A and 0.250A but I don't understand how to solve for part c.

please help me

Answer 1

Concepts and reason

The concepts required to solve this problem are the Kirchhoff’s circuit laws and the Ohm’s law.

Initially, calculate the current drawn by the good battery alone through the batteries using the ohm’s law. Later, calculate the current that passes through the different resistors by applying the Kirchhoff’s law. Finally, apply the loop and solve for the current.

Fundamentals

The expression of the current according to the Ohm’s law is,

Here, I is the current, V is the voltage, and R is the resistance.

The equivalent resistance when two resistors are connected in series is,

Here, and are the two-resistance connected in series.

The Kirchhoff’s voltage law states that the sum of all the voltages around a closed loop in an electrical circuit will be equal to the zero, that is,

Here, V is the voltage.

The Kirchhoff’s current law states that at any junction, the sum of the currents that flow into that node will be equal to the sum of currents that flow out of that node.

(a)

The following is the circuit diagram showing the loops ABEF and CBDE:

$\B В C A 30.0 20.00 10.0 V Ω 5.00 V F E Figure 1: Circuit diagram showing the loop ABEF and CBDE. $

Apply the Kirchhoff’s loop law to the loop CBED as follows:

Therefore,

…… (2)

Subtract equation (2) from the equation (1) as follows:

Solve for .

Now, substitute the value of I the equation (2) and solve for as follows:

On adding the currents and , the current across the is calculated as follows:

Substitute for and for in the equation .

(b)

Subtract equation (2) from the equation (1) as follows:

The equation obtained is as follows:

Solve for .

(c)

Refer figure 1.

The current through the 10.0 V battery is equal to . Therefore,

Ans: Part a

The current passing through the resistor is equal to 0.333 A.

Part b

The current passing through the resistor is equal to 0.25 A.

Part c

The current through the 10.0 V battery is equal to 0.583 A.