Find the voltage (emf).

Electricity and Magnetism Level pending

You have a circuit as presented in the diagram.

The ammeter reads $\SI{0}{\ampere}.$ Find the voltage/emf (in volts) of the battery $E$ .

The answer is 8.

2 solutions

Chew-Seong Cheong
Feb 18, 2017

When the ammeter read 0 A, this means that there is no current flowing through 3.0 $\Omega$ resistor. Hence, there is no voltage drop across the 3.0 $\Omega$ resistor and voltage at node A, $V_{AB} = E$ . Since there is no current through the ammeter, the ammeter can be replaced with an open circuit. Then the remaining circuit is a voltage divisor and:

$\begin{aligned} V_{AB} & = \frac {4}{2+4} \times 12 \\ E & = \boxed{8}V \end{aligned}$

Can you explain how you arrive at the expression of $V_{AB}$ ? (This will allow others who can not solve the problem to understand your solution fully).

Rohit Gupta - 4 years, 3 months ago

Sorry, I forgot the diagram.

Chew-Seong Cheong - 4 years, 3 months ago

Thanks, the diagram is really helpful here. It shows how the two resistors come in series. Then we can easily get $V_{\text{AB}} = i \times 4 \Omega$ , where $i = \dfrac{12}{2+4}$ . This will give us $V_{\text{AB}} = 8V$ which is also equals to $E$ as the voltage drop across $3 \Omega$ resistor is zero.

Rohit Gupta - 4 years, 3 months ago

Tom Engelsman
Feb 17, 2017

Let's apply KCL at the LHS middle junction (taking the RHS middle junction to be ground (0V)). If the ammeter reads zero amps, then the LHS middle junction is at the same potential of $E$ volts. We now require:

$\frac{E - E}{3\Omega} + \frac{E - 12}{2\Omega} = \frac{0 - E}{4\Omega} \Rightarrow 2E - 24 = -E \Rightarrow E = \boxed{8V}.$

Hey, I feel you are missing a gap here, can you elaborate further how have you arrived at this equation. Maybe a diagram will help.

Rohit Gupta - 4 years, 3 months ago

Find the voltage (emf).

The answer is 8.

2 solutions

0 pending reports