Source: High school physics (Chinese)
Problem Sets:
Problem
In the circuit shown, the voltage between points $a$ and $b$ is $U_{ab} = 9.0$ V. The network has two parallel branches between $a$ and $b$. The first branch is a $2$ k$\Omega$ resistor in series with a $10$ k$\Omega$ resistor. The second branch is a $1$ k$\Omega$ resistor in series with a $5$ k$\Omega$ resistor. (In the figure, "k" denotes kilohms.)
- Find the current through each resistor.
- Find the voltage across each resistor.
Currents: $I_{2k} = I_{10k} = 0.75$ mA; $I_{1k} = I_{5k} = 1.5$ mA.\ Voltages: $U_{2k} = U_{1k} = 1.5$ V; $U_{10k} = U_{5k} = 7.5$ V.
Each branch is driven by the full $U_{ab} = 9$ V.
Branch 1 (total resistance $2 + 10 = 12$ k$\Omega$):
$$I_{2k} = I_{10k} = \frac{9 \text{ V}}{12 \text{ k}\Omega} = 0.75 \text{ mA}.$$ $U_{2k} = 0.75 \times 2 = 1.5$ V; $U_{10k} = 0.75 \times 10 = 7.5$ V.Branch 2 (total resistance $1 + 5 = 6$ k$\Omega$):
$$I_{1k} = I_{5k} = \frac{9 \text{ V}}{6 \text{ k}\Omega} = 1.5 \text{ mA}.$$ $U_{1k} = 1.5 \times 1 = 1.5$ V; $U_{5k} = 1.5 \times 5 = 7.5$ V.