Compound Direct Current Circuits - College Physics

A student builds a circuit to measure an unknown resistance $R_x$ using a Wheatstone bridge configuration. The bridge uses known resistors: $R_1 = 100\,\Omega$ , $R_2 = 150\,\Omega$ , and $R_3 = 200\,\Omega$ . When the bridge is balanced (zero current through the galvanometer), what is $R_x$ , and what happens to the bridge balance if $R_1$ is increased to $120\,\Omega$ ?
Answer: $R_x = 300\,\Omega$ ; current flows from junction A to junction B through the galvanometer For a balanced Wheatstone bridge, $\frac{R_1}{R_2} = \frac{R_3}{R_x}$ . Substituting: $\frac{100}{150} = \frac{200}{R_x}$ , which gives $R_x = \frac{200 \times 150}{100} = 300\,\Omega$ . When $R_1$ increases to $120\,\Omega$ , the ratio $\frac{R_1}{R_2} = \frac{120}{150} = 0.8$ becomes greater than $\frac{R_3}{R_x} = \frac{200}{300} = 0.667$ . This means the voltage at junction A (between $R_1$ and $R_2$ ) becomes higher than the voltage at junction B (between $R_3$ and $R_x$ ). Therefore, current flows from the higher potential junction A to the lower potential junction B through the galvanometer.
A complex circuit network contains 5 nodes and multiple resistors. Using Kirchhoff's current law (KCL) at each node, a system of equations is set up. If the currents $I_1 = 3\text{ A}$ , $I_2 = -1\text{ A}$ , $I_3 = 2\text{ A}$ , and $I_4 = -2\text{ A}$ are known, and the equation for node E is $I_3 + I_5 - I_6 + I_7 = 0$ , what additional constraint is needed to solve for $I_5$ , $I_6$ , and $I_7$ ?
Answer: Both KVL equations and resistance values, since KCL alone provides insufficient constraints The single KCL equation $I_3 + I_5 - I_6 + I_7 = 0$ provides only one constraint for three unknowns ( $I_5$ , $I_6$ , $I_7$ ). With $I_3 = 2\text{ A}$ , we get $2 + I_5 - I_6 + I_7 = 0$ , or $I_5 - I_6 + I_7 = -2$ . This is one equation with three unknowns, which is insufficient. We need two additional independent equations. These must come from Kirchhoff's voltage law (KVL) around loops that contain these current-carrying branches, combined with Ohm's law ( $V = IR$ ) which requires knowing the resistance values. KCL equations alone cannot provide the additional constraints needed because they only relate currents at nodes, not the voltage drops that determine the current magnitudes through specific resistances.
Two identical batteries, each with EMF $12\text{ V}$ and internal resistance $2\,\Omega$ , are connected in parallel to a load resistor $R_L$ . If the power delivered to the load is maximum, what is the efficiency of power transfer from the batteries to the load?
Answer: $50\%$ Two identical batteries in parallel have equivalent EMF $12\text{ V}$ and equivalent internal resistance $2\,\Omega/2 = 1\,\Omega$ . Maximum power transfer occurs when the load resistance equals the internal resistance, so $R_L = 1\,\Omega$ . The total circuit resistance is $1 + 1 = 2\,\Omega$ , giving current $I = 12/2 = 6\text{ A}$ . Power delivered to load: $P_L = I^2 R_L = 36 \times 1 = 36\text{ W}$ . Power lost in internal resistance: $P_{internal} = I^2 \times 1 = 36\text{ W}$ . Total power: $P_{total} = 36 + 36 = 72\text{ W}$ . Efficiency = $P_L/P_{total} = 36/72 = 50\%$ . This is a general result: maximum power transfer always occurs at 50% efficiency.
A circuit contains three identical resistors, each with resistance $R$ , connected to form a triangle. A battery with EMF $\mathcal{E}$ and internal resistance $r$ is connected between two vertices of the triangle. What is the current delivered by the battery?
Answer: $\frac{3\mathcal{E}}{2R + 3r}$ The three resistors form a triangle, and the battery is connected between two vertices. From the battery's perspective, one resistor ( $R$ ) is in parallel with the series combination of the other two resistors ( $2R$ ). The equivalent resistance of this parallel combination is $\frac{R \cdot 2R}{R + 2R} = \frac{2R^2}{3R} = \frac{2R}{3}$ . The total resistance in the circuit is the internal resistance plus this equivalent resistance: $r + \frac{2R}{3}$ . Therefore, the current is $I = \frac{\mathcal{E}}{r + \frac{2R}{3}} = \frac{3\mathcal{E}}{3r + 2R} = \frac{3\mathcal{E}}{2R + 3r}$ .

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