Question

The ideal battery in the given figure (a) has emf $E = 10.0 V$ . Plot $1$ in the figure (b) gives the electric potential difference $V$ that can appear across resistor 1 versus the current $i$ in that resistor when the resistor is individually tested by putting a variable potential across it. The scale of the $V$ axis is set by $V_{s} = 18.0 V$ , and the scale of the $i$ axis is set by $i_{s} = 3.00 mA$ . Plots $2$ and $3$ are similar plots for resistors $2$ and $3$ , respectively, when they are individually tested by putting a variable potential across them. What is the current in resistor $2$ in the circuit of figure a?

$(a)$

$(b)$

Accepted Answer

The equivalent resistance of circuit.

Here $R_{1} a n d R_{2}$ parallel

$V_{s} = 18.0 V$

The resultant of $R_{1} a n d R_{2}$ is series with $R$ ,

$R = \frac{R_{1} R_{2}}{R_{1} + R_{2}} + R_{3}$

$18 V = V_{s}$

Voltage on $Y$ -axis

$1 Unit = 2 V$

$V_{s} = 18 V$

Current on $X$ -aixs

$E = 10.0 V$

The slope of voltage and current graph gives resistance

The slope of curve(1), $R_{1} = \frac{18 V}{3 mA} = 6 kΩ$

The slope of curve $(2) R_{2} = \frac{12 V}{3 mA} = 4 kΩ$

The slope of curve $(3) R_{3} = \frac{6 V}{3 mA} = 2 kΩ$

The equivalent resistance of circuit

$R = \frac{R_{1} R_{2}}{R_{1} + R_{2}} + R_{3}$

$= \frac{(6 kΩ) (4 kΩ)}{(6 kΩ) + (4 kΩ)} + (2 kΩ)$

$R = 4.4 kΩ$

The current through the battery

$i = \frac{ε}{r}$

Since, $ε = 10 V a n d R = 4.4 kΩ$

$i = \frac{10 V}{4.4 kΩ}$

$= 2.27 mA$

The voltage drop across the resistor $(R_{3})$

$V_{3} = i R_{3}$

$V_{3} = (2.27 mA) (2 kΩ)$

$V_{3} = 4.54 V$

The remaining voltage

$V_{2} = ε - V_{3}$

$= 10 V - 4.54 V$

$= 5.46 V$

This $V_{2} = 5.46 V$ is applied across $R_{1} a n d R_{2}$ is same.

The current through resistor $(R_{2})$ is

$i_{2} = \frac{V_{2}}{R_{2}}$

$= \frac{5.46}{4 K}$

$= 1.365 mA .$

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