Irodov 3.183: The Ideal Source Hides a Resistor – Amit & Deepali Sinha Foundation

Problem Irodov 3.183

A rectangular circuit has a battery of emf $\mathcal{E}_0$ on its left side, positive terminal at the bottom, and a battery of emf $\mathcal{E}$ on its right side, positive terminal at the top. Along the top wire, from left to right, a resistor $R_2$ lies between two junctions. From the left junction a resistor $R_1$ , and from the right junction a resistor $R_3$ , run down to the bottom wire. A resistor $R$ sits in the bottom wire between the foot of $R_3$ and the battery $\mathcal{E}$ . The internal resistances of the batteries are negligible. Find the current through $R$ .

The circuit. E_0 has its long (+) plate at the bottom, E its long (+) plate at the top. — The circuit. $\mathcal{E}_0$ has its long ( $+$ ) plate at the bottom, $\mathcal{E}$ its long ( $+$ ) plate at the top.

Given

$\mathcal{E}_0$
$\mathcal{E}$
$R_1$ , $R_2$ , $R_3$ , $R$
internal resistances negligible

Find

the current

I

through

R

Solution

Step 1The fixed potentials

Name the upper junctions $P$ (left) and $Q$ (right), and take the bottom wire from $\mathcal{E}_0$ to the foot of $R_3$ as the zero of potential, $O$ . Let $I$ be the current through $R$ from left to right, as drawn.

The junctions P and Q, the bottom wire O at zero potential, and the current I through R, taken from left to right. — The junctions $P$ and $Q$ , the bottom wire $O$ at zero potential, and the current $I$ through $R$ , taken from left to right.

The battery $\mathcal{E}_0$ joins $O$ (at its $+$ plate) directly to $P$ (at its $-$ plate), and it has no internal resistance, so it fixes $P$ whatever flows through it:

\varphi_P = -\mathcal{E}_0 \tag{1}

Walking from $O$ through $R$ with the current, the potential falls by $IR$ ; then across $\mathcal{E}$ , from $-$ to $+$ , it rises by $\mathcal{E}$ and reaches $Q$ :

\varphi_Q = \mathcal{E} - IR \tag{2}

Step 2The junction rule at

Q

Three branches meet at $Q$ . The current $I$ arrives through $\mathcal{E}$ ; the currents $\varphi_Q/R_3$ through $R_3$ and $(\varphi_Q - \varphi_P)/R_2$ through $R_2$ leave it. Their balance:

I = \frac{\varphi_Q}{R_3} + \frac{\varphi_Q - \varphi_P}{R_2}

Substituting (1) and (2):

I = \frac{\mathcal{E} - IR}{R_3} + \frac{\mathcal{E} + \mathcal{E}_0 - IR}{R_2}

Step 3Solving for the current

Multiplying through by $R_2R_3$ :

IR_2R_3 = (\mathcal{E} - IR)R_2 + (\mathcal{E} + \mathcal{E}_0 - IR)R_3

Collecting the terms in $I$ on the left:

I(R_2R_3 + RR_2 + RR_3) = \mathcal{E}(R_2 + R_3) + \mathcal{E}_0R_3

Dividing by the bracket:

I = \frac{\mathcal{E}(R_2 + R_3) + \mathcal{E}_0R_3}{R(R_2 + R_3) + R_2R_3}

(3)

For positive emfs it is positive: the current passes $R$ from left to right, as drawn, and rises through $\mathcal{E}$ .

Check

With $\mathcal{E}_0 = 0$ the left battery is a plain wire, $P$ joins $O$ , and $R_2$ and $R_3$ are in parallel between $Q$ and $O$ ; the formula gives $I = \mathcal{E}/(R + R_2R_3/(R_2 + R_3))$ , the current of $\mathcal{E}$ through $R$ in series with that pair, as it must. With $\mathcal{E} = 0$ , the current $\mathcal{E}_0R_3/(R(R_2 + R_3) + R_2R_3)$ is what $\mathcal{E}_0$ alone sends through $R$ , which is in parallel with $R_3$ and in series with $R_2$ . Units: amperes in both cases.

Remark

The answer has no $R_1$ in it. A resistor connected straight across an ideal source only draws the extra current $\mathcal{E}_0/R_1$ from that source and fixes nothing that was not already fixed. The answer is also the sum of the two special cases in the check: in a linear network the current due to several sources is the sum of the currents each would drive alone, the superposition principle.