For the DC circuit below, \$E_1=15V\$, \$R_1=R_2=4k\Omega\$, \$R_3=1k\Omega\$ and \$R_4+R_5=3k\Omega\$. When switch \$P\$ is open, the voltmeter (with internal resistance \$R_v=10k\Omega\$) reads the voltage \$U_{21}=15V\$. Find the current \$I\$ through resistor \$R_3\$ when switch \$P\$ is closed, if, in that case (closed switch), the voltmeter reads \$20V\$.
What I did to solve this is find \$I_g\$ and \$R_6\$ in the first state (when \$P\$ is open) and then find the current \$I\$ using superposition of currents from sources \$E_1\$, \$E_2\$ and \$I_g\$) in the second state (when \$P\$ is closed). But I got a wrong result.
Then I got really confused by the voltmeter readings. If the voltmeter shows \$15V\$ when switch \$P\$ is open, that means that \$-I_gR_6=15V\$, but in the second state, when \$P\$ is closed, we have \$-I_gR_6=20V\$. How can that be?
Best Answer
"Then I got really confused by the voltmeter readings. If the voltmeter shows 15V when switch P is open, that means that −IgR6=15V, but in the second state, when P is closed, we have −IgR6=20V. How can that be?"
You are assuming the voltage across Ig is zero. It could be but is not required to be.
A resistor in series with a current source never changes anything as the voltage across the current source will always adjust itself to compensate.
I would suggest redrawing the circuit with a resistor in place of the Voltmeter (10k).
short out R4 and replace R5 with a 3k resistor.
Personally I would also draw the circuit up the other way with the positive side of E1 and E2 upwards. These will help in understanding the circuit better.