Irodov 3.178: Two Loads on a Real Source – Amit & Deepali Sinha Foundation

Problem Irodov 3.178

A source of emf $\mathcal{E} = 5.0$ V and internal resistance $R = 0.10~\Omega$ feeds two resistors, $R_1 = 4.0~\Omega$ and $R_2 = 6.0~\Omega$ , connected in parallel across its terminals. Find the currents through $R_1$ and through $R_2$ .

The source, its long (+) plate on top and its internal resistance R written beside it, with R_1 and R_2 in parallel across it. — The source, its long ( $+$ ) plate on top and its internal resistance $R$ written beside it, with $R_1$ and $R_2$ in parallel across it.

Given

$\mathcal{E} = 5.0$ V
$R = 0.10~\Omega$ (internal)
$R_1 = 4.0~\Omega$
$R_2 = 6.0~\Omega$

Find

I_1

I_2

Solution

Step 1The current drawn from the source

The source drives $I$ up through itself and out of its $+$ terminal; at the top junction it divides into $I_1$ through $R_1$ and $I_2$ through $R_2$ , as drawn.

The source current I and its two parts, I_1 through R_1 and I_2 through R_2. — The source current $I$ and its two parts, $I_1$ through $R_1$ and $I_2$ through $R_2$ .

The two resistors in parallel act as one of resistance

R_{12} = \frac{R_1R_2}{R_1 + R_2}

which is in series with the internal resistance $R$ . The loop rule then gives the source current:

I = \frac{\mathcal{E}}{R + R_{12}} = \frac{\mathcal{E}(R_1 + R_2)}{R(R_1 + R_2) + R_1R_2}

Step 2The voltage across the pair

Both resistors have the same voltage $V$ across them, the voltage of the source’s terminals. It is $I$ times the combined resistance:

V = IR_{12} = \frac{\mathcal{E}R_1R_2}{RR_1 + R_1R_2 + R_2R}

(1)

Step 3The two branch currents

Each branch current is the common voltage (1) divided by that branch’s resistance. With the numbers, the common denominator is

\begin{aligned} RR_1 + R_1R_2 + R_2R &= (0.10)(4.0) + (4.0)(6.0) + (6.0)(0.10)\\ &= 25~\Omega^2 \end{aligned}

so that

\begin{aligned} I_1 &= \frac{(5.0)(6.0)}{25}~\text{A}\\[2pt] I_2 &= \frac{(5.0)(4.0)}{25}~\text{A} \end{aligned}

which gives

\begin{gathered} I_1 = \frac{\mathcal{E}R_2}{RR_1 + R_1R_2 + R_2R} = 1.2\text{ A}\\[4pt] I_2 = \frac{\mathcal{E}R_1}{RR_1 + R_1R_2 + R_2R} = 0.80\text{ A} \end{gathered}

(2)

Both flow downward through their resistors, from the top wire to the bottom one.

Check

The two currents must add up to the source current: $I_1 + I_2 = 2.0$ A, and $\mathcal{E}/(R + R_{12}) = 5.0/(0.10 + 2.4) = 2.0$ A. They must also give one voltage: $I_1R_1 = I_2R_2 = 4.8$ V. Limit: as $R \to 0$ the currents tend to $\mathcal{E}/R_1 = 1.25$ A and $\mathcal{E}/R_2 = 0.83$ A, each resistor then seeing the full emf.

Remark

The ratio $I_1/I_2 = R_2/R_1$ does not depend on the source at all: a parallel pair always splits its current inversely to the resistances. The internal resistance only sets how much there is to split. Here it is small against the load ( $0.10~\Omega$ against $2.4~\Omega$ ), so the terminal voltage $4.8$ V is only $4$ per cent below the emf.