Ohm's Law

In the circuit of Figure, the source has EMF $\varepsilon = 4$ V and internal resistance $r$. The external network has $R_1 = 4$ $\Omega$ and the series pair $R_2 + R_3 = 2 + 10 = 12$ $\Omega$ connected as two parallel branches between two nodes, and the parallel combination is in series with $R_4 = 6$ $\Omega$. An ideal ammeter is in series with the $R_1$ branch and reads $0.3$ A. Voltmeter $V_1$ is connected across $R_3$ in the middle branch; voltmeter $V_2$ is connected across the source terminals.

The total resistance of a transmission line is $1.0$ $\Omega$, and the transmitted power is $100$ kW.

In the bridge network of Figure, four nodes $A$, $B$, $C$, $D$ are arranged as the vertices of a diamond. The outer resistors are $R_{AC} = 10$ $\Omega$ (between $A$ and $C$), $R_{CB} = 5$ $\Omega$ (between $C$ and $B$), $R_{AD} = 20$ $\Omega$ (between $A$ and $D$), $R_{DB} = 10$ $\Omega$ (between $D$ and $B$), and a bridge resistor $R_{CD} = 100$ $\Omega$ across the diagonal between $C$ and $D$. Find the equivalent resistance when the following pairs of terminals are connected to a circuit:

In the circuit of Fig. 13.47, four nodes are arranged as a rectangle: $A$ (top-left), $C$ (top-right), $B$ (bottom-left), $D$ (bottom-right). The resistor $R_1$ is on the left side between $A$ and $B$; $R_2$ is on the top between $A$ and $C$; $R_3$ is the diagonal between $B$ and $C$; $R_4$ is on the bottom between $B$ and $D$. A switch $S$ is on the right side between $C$ and $D$. All four resistors are equal: $R_1 = R_2 = R_3 = R_4 = R$.

In each of the circuits of Figure, find the labeled physical quantities.

Circuit (a): A $1$ k$\Omega$ resistor and a $5$ k$\Omega$ resistor are connected in parallel between two nodes. A current of $1$ mA flows through the $1$ k$\Omega$ branch. The current $I$ enters the parallel combination from the left, and $I_1$ denotes the current through the $5$ k$\Omega$ branch.

Circuit (b): A $5$ k$\Omega$ resistor and a $1$ k$\Omega$ resistor are connected in series. The voltage across the $5$ k$\Omega$ resistor is $10$ V, and the voltage across the $1$ k$\Omega$ resistor is $U$. The current through the series circuit is $I$.

Circuit (c): Resistor $R$ is in parallel with a $3$ $\Omega$ resistor, and this parallel combination is in series with a $4$ $\Omega$ resistor. The supply voltage is $12$ V and the current through the $4$ $\Omega$ resistor is $2$ A.

Circuit (d): Three resistors of $1.0$ k$\Omega$, $2.0$ k$\Omega$, $3.0$ k$\Omega$ are connected in parallel and carry currents $I_1$, $I_2$, $I_3$ respectively. The total current entering the combination is $1.0$ mA.

The resistor network of Figure has terminals $a$ and $b$ on the left side; its right end is open. Both the top and bottom rails consist of six $2$ $\Omega$ resistors in series. Three vertical rungs connect the rails: a $6$ $\Omega$ rung after the first resistor on each rail, a second $6$ $\Omega$ rung after the third resistor, and a $4$ $\Omega$ rung after the fifth resistor. The two $2$ $\Omega$ resistors after the last rung (one on each rail) terminate at the open right end.

In the circuit of Figure, $R_1 = 10$ $\Omega$, $R_2 = 4$ $\Omega$, $R_3 = 6$ $\Omega$, $R_4 = 3$ $\Omega$, and the source voltage is $U = 2.4$ V. The circuit topology is: the source connects between top node $T$ (positive) and ground node $G$ (negative). $R_1$ goes from $T$ down to a middle node $M$; $R_4$ goes from $M$ down to $G$. $R_2$ goes from $T$ down to node $N$, and $R_3$ goes horizontally from $M$ to $N$. Terminal $a$ is at node $N$; terminal $b$ is at $G$ (on the bottom rail).

A battery with EMF $\mathcal{E}$ and internal resistance $r$ is connected to a circuit consisting of two parallel branches: branch 1 contains resistor $R_1$, and branch 2 contains resistor $R_2$ in series with a slide rheostat $R_3$. Ammeter $A_1$ is in series with the battery (measuring the total current), and ammeter $A_2$ is in the $R_2$-$R_3$ branch. Voltmeter $V$ reads the battery terminal voltage, voltmeter $V_1$ reads the voltage across $R_1$, and voltmeter $V_2$ reads the voltage across $R_3$.

A slide rheostat of total resistance $R = 3 \text{ k}\Omega$ has its two end terminals $A$ (top) and $C$ (bottom) connected to a $U = 10 \text{ V}$ source. A load resistor $R_3 = 3 \text{ k}\Omega$ is connected between the sliding contact $B$ and the lower terminal $C$, so that $R_3$ is in parallel with the lower portion $BC$ of the rheostat.

The circuit of Figure has four resistors in series between terminals $A$ and $B$, in the order $R_1 \to R_2 \to R_3 \to R_4$. The values are $R_1 = R_2 = 4$ $\Omega$, $R_3 = R_4 = 2$ $\Omega$, and the applied voltage is $U_{AB} = 6$ V. Two ideal ammeters (zero internal resistance) provide bypass paths: ammeter $A_1$ is connected from terminal $A$ to the node between $R_2$ and $R_3$ (so $A_1$ short-circuits the series pair $R_1$-$R_2$); ammeter $A_2$ is connected from the node between $R_1$ and $R_2$ to the node between $R_3$ and $R_4$ (so $A_2$ short-circuits the series pair $R_2$-$R_3$).