Three batteries have their negative terminals connected together.

Find the voltage read by an ideal voltmeter.

###### Solution.

We assume that the potential of the point A is zero. Then, the potential of the point D is 3V, the potential of the point E is 6V and the potential of the point C is 9V. We can denote the potential of the point B as V_{B}.

The voltage V_{DB} on the resistor is 3 – V_{B} and the current equals (3 – V_{B})/100

and the voltage V_{EB} on the resistor is 6 – V_{B} and the current equals (6 – V_{B})/100.

Also, the voltage V_{CB} on the resistor is 9 – V_{B} and the current equals (9 – V_{B})/200.

According to Kirchhoff’s current law, at any instant, the algebraic sum of the currents entering a node in the circuit is zero.

Thereby,

(3 – V_{B})/100 + (6 – V_{B})/100 + (9 – V_{B})/200 = 0

6 – 2V_{B} + 12 – 2V_{B} + 9 – V_{B} = 0

5V_{B} = 27

V_{B} = 5.4V

Finally, the voltage read by an ideal voltmeter is 5.4V.