Name: RC Circuit and Charging Capacitor
Uploaded: 2019-07-19
Duration: 13 min 36 s
Description: In this video, we will discuss the physics of charging the capacitor connected in a circuit and derive the expression and relation between the parameters inv...

Question

In given circuit determine

$(a)$ Equivalent resistance (Including internal resistance).

$(b)$ Current $i, i_{1}, i_{2}$ and $i_{3}$ .

$(c)$ The potential difference across battery and each resistance.

$(d)$ The rate at which the chemical energy of the cell is consumed.

$(e)$ The rate at which heat is generated inside the battery.

$(f)$ Electric power output.

$(g)$ Which resistance consumes maximum power?

$(h)$ Power dissipated across $4 Ω$ resistance.

Accepted Answer

(a) In the given circuit, all three resistances $(a)$ , $4 Ω$ and $8 Ω$ are in parallel combination, so their equivalent resistance will be,

$\frac{1}{R} = \frac{1}{8} + \frac{1}{4} + \frac{1}{8}$

$R = 2 Ω$ ,

This $2 Ω$ resistance will be in series with the internal resistance, so the final equivalent resistance of the circuit will be,

$R_{eq} = R + r = 2 + 1 = 3 Ω$

(b) The current drawn from the battery will be,

$i = \frac{ε}{R_{eq}} = \frac{6}{3} = 2 A$

From Kirchhoff's law, the potential across all three resistances will be given by,

$V = 6 - r i = 6 - 1 \times 2 = 4 V$

All resistances are in parallel combination, so the potential across them will be same and the current corresponding to the resistance is given by,

$i_{1} = \frac{V}{8} = \frac{4}{8} = \frac{1}{2} A$

$i_{2} = \frac{V}{4} = \frac{4}{4} = 1 A$

$i_{3} = \frac{V}{8} = \frac{4}{8} = \frac{1}{2} A$

(c) As all resistances are in parallel combination, so potential will be $5 V$ across each resistance.

(d) Rate of chemical energy consumed by the cell is given by,

$P_{cell} = ε i = 6 \times 2 = 12 W$

(e) Rate of heat generated in the battery is given by,

$H = i^{2} r = {(2)}^{2} \times 1 = 4 W$

(f) Some of chemical energy is lost due to heating, so, the electric power output will be,

$P_{out} = P_{cell} - H = 12 - 4 = 8 W$

(g) The power consume across the resistances connected in parallel is given by,

$P = \frac{V^{2}}{R}$

$P \propto \frac{1}{R}$ ; means the least resistance will consume more power, so $4 Ω$ will consume maximum power.

(h) The power across $4 Ω$ resistance will be,

$P_{4} = \frac{V^{2}}{4} = \frac{4^{2}}{4} = 4 W$ .

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