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 the figure each cell has an emf of $1.5 V$ and internal resistance of $0.40 Ω$ . Calculate:

(i) current $I$

(ii) current in the $36 Ω$ resistor

(iii) potential difference across $A$ and $B$ .

Accepted Answer

First we need to find out the net equivalent potential and resistance provided by the series-parallel combination of cell in the given circuit. The four resistances are connected in series with emf $1.5 V$ and internal resistance $0.40 Ω$ , so their equivalent potential will be,

$ε_{1} = 4 \times 1.5 = 6 V$

And their internal resistance will be,

$r_{1} = 4 \times 0.40 = 1.6 Ω$

They are connected in parallel with the same set of the cells, so the combination of cell will give equivalent emf,

$ε = \frac{\frac{6}{1.6} + \frac{6}{1.6}}{\frac{1}{1.6} + \frac{1}{1.6}} = 6 V$

The equivalent resistance will be,

$r = \frac{1.6 \times 1.6}{1.6 + 1.6} = 0.8 Ω$

The overall circuit will be given by,

(i) All three resistance of $12 Ω$ , $18 Ω$ and $36 Ω$ are in parallel combination, so their equivalent resistance will be,

$\frac{1}{R} = \frac{1}{12} + \frac{1}{18} + \frac{1}{36}$

$R = 6 Ω$ .

Apply Kirchhoff's loop rule in the given circuit,

$6 I + 1.5 + 0.40 I + 1.8 I + 0.8 I - 6 = 0$

$I = 0.50 A$ .

(ii) The $12 Ω$ , $18 Ω$ and $36 Ω$ are connected in parallel, and current through their equivalent combination is $0.50 A$ , so the potential across these resistances will be same and given by,

$V_{A D} = 6 \times I = 6 \times 0.5 = 3 V$

Hence, the current through the $36 Ω$ resistance will be,

$I_{36 Ω} = \frac{3}{36} = 0.0833 A$ .

(iii) The potential difference across $V_{A B}$ can be given by,

$V_{A B} = 1.5 + 0.4 \times 0.5 = 1.7 V$ .

In the figure each cell has an emf of $1.5 V$ and internal resistance of $0.40 Ω$ . Calculate:

(i) current $I$

(ii) current in the $36 Ω$ resistor

(iii) potential difference across $A$ and $B$ .

Important Questions on Current Electricity

Learn and Prepare for any exam you want !

In the figure each cell has an emf of 1.5 V and internal resistance of 0.40 Ω. Calculate: (i) current I (ii) current in the 36 Ω resistor (iii) potential difference across A and B.

Important Questions on Current Electricity

Learn and Prepare for any exam you want !

In the figure each cell has an emf of $1.5 V$ and internal resistance of $0.40 Ω$ . Calculate:

(i) current $I$

(ii) current in the $36 Ω$ resistor

(iii) potential difference across $A$ and $B$ .