Chapter 23.2, Problem 30SSC

To determine

the current through the bulb 2.

Answer to Problem 30SSC

The current flowing through the bulb 2 is 0.6 A.

Explanation of Solution

Given:

Four resistors are connected.

The circuit diagram is

  

  $I_3=1.7\text{A}$

  $I_1=1.1\text{A}$

Formula used:

Kirchhoff’s law can be expressed as

  $I_3=I_2+I_1$  ...... (1)

Here, $I_1$ and $I_2$ are the current moving away from the junction $q$ in the circuit and $I_3$ is the current moving towards the junction $q$ in the circuit.

Calculation:

Kirchhoff’s first rule states that the sum of incoming current towards a junction in a circuit is equal to the sum of outgoing current away from that junction.

The circuit diagram is as shown below

  

In the above circuit, 1,2 and 3 are the three bulbs.

Rearrange the equation (1) for $I_2$ ,

  $I_2=I_3-I_1$  ......(2)

Substitute the values in equation (2)

  $\begin{array}{l}{I}_2=\left(1.7-1.1\right)\text{A}\\ \boxed{I_2=0.6\text{A}}\end{array}$

Conclusion:

Hence, the current flowing through the bulb 2 is 0.6 A.