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 following circuit potential at point $‘ A ’$ is zero then determine

(a) Potential at each point.

(b) Potential difference across each resistance.

(c) Identify the batteries which act as a source.

(d) Current in each battery.

(e) Which resistance consumes maximum power?

(f) Which battery consume or gives maximum power?

Accepted Answer

(a) Given that potential at $A$ $= 0$ and potential difference between $E$ and $A$ is $10 V$ which is given as, $V_{E} - V_{A} = 10$

$\Rightarrow V_{E} = 10 V$

It is clear from the given circuit diagram that points $E, F, G, H$ are connected to one another, so the potential at these points will be equal i.e; $10 V .$

Similarly, points $A, B, C, D$ are connected to one another, so the potential at these points will be same i.e; $0 V .$

As $10 V$ supply is in series with $5 V$ supply between points $F$ and $B$ , So, the potential at $I = 10 + 5 = 15 V .$

Now, potential at $J = 15 V .$

Similarly, potential at $K$ will be the sum of potentials at $H$ and the $5 V$ source (as connected in series) i.e; $V_{K} = 10 + 5 = 15 V .$

(b) Potential difference across resistor connected between points $F$ and $B$ is $V_{I} - V_{B} = 15 - 0 = 15 V,$

Potential difference across resistor connected between points $G$ and $C$ is $V_{J} - V_{G} = 15 - 10 = 5 V$ ,

Potential difference across resistor connected between points $H$ and $D$ is $V_{K} - V_{D} = 15 - 0 = 15 V .$

(c) From the given circuit diagram we can see that the $10 V$ source battery is in series with the two $5 V$ batteries which are in parallel also the $15 V$ battery is in series with the other two $5 V$ batteries, So, all the batteries act as source.

(d) Current in the battery between points $F$ and $B$ is $= \frac{15}{1} = 15 A,$

Current in the battery between points $G$ and $C$ is $= \frac{5}{2} = 2.5 A$ ,

Current in the battery between points $H$ and $D$ is $= \frac{15}{3} = 5 A,$

Now, for current (say $I$ ) in the battery of $10 V$ , lets reduce all the other batteries into current source as shown in the diagram below and applying $K C L$ at point $O$ then, we get,

$I + 2.5 - 15 - 5 = 0$

$\Rightarrow I = 20 - 2.5 = 17.5 A .$

(e) Power in each resistor is given as,

For $1 Ω$ resistor, $P = \frac{V^{2}}{R} = \frac{15^{2}}{1} = 225 W,$

For $2 Ω$ resistor, $P = \frac{V^{2}}{R} = \frac{5^{2}}{2} = 12.5 W,$

For $3 Ω$ resistor, $P = \frac{V^{2}}{R} = \frac{15^{2}}{3} = \frac{225}{3} = 75 W .$

Hence, we can see $1 Ω$ resistor consumes maximum power.

(f) Power in each battery is given as,

For $10 V$ battery, $P = V I = 10 \times 17.5 = 175 W,$

For $5 V$ battery, $P = V I = 5 \times 15 = 75 W,$

For $15 V$ battery, $P = V I = 15 \times 2.5 = 37.5 W,$

For last $5 V$ battery at the extreme right side of the circuit, $P = 5 \times 5 = 25 W .$

So, it is clear that $10 V$ battery uses maximum power.

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