Question

In the electrical circuit shown, points B and C are earthed. Find the potential of point A and D


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Answer 1

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First of all calculate the current flowing through the circuit i.e, I

           $V=IR\\I=\frac{V}{R}$

But equivalent resistance is given by :

Assuming 1Ω Resistance to be internal resistance

$E=I(R+r)$

where E is EMF of cell

I is current flowing through the circuit

R is external resistance

r is internal resistance

$R \:can\:be\:calculated\:as \\since\: 3\:external\:resistors\:are\:in\:series\:hence :\\R_{s}=R_{1}+R_{2}+R_{3}\\R_{s}=4+10+6\\R=20 ohm$

$E=I(20+1)\\10=21\timesI\\I=\frac{10}{21}$

FROM KIRCHOFF'S LAW

But Vb and Vc are earthed hence these must be equal to 0

$V_{A}-4I+0-10I+0-6I=V_{D\\}\\V_{A}-V_{D}=20I\\V_{A}-V{D}=\frac{40}{21}$

Answer 2

Answer:

The potential of points A and D will be $\frac{40}{21}$.

Explanation:

First, figure out how much current is passing through the circuit, i.e, I

V = IR

I = $\frac{V}{R}$

But comparable resistance is provided by: presuming 1Ω Internal resistance is what resistance is.

E = I(R + r) where E is EMF of the cell, I is current flowing through the circuit and R is external resistance.

Now let's calculate value of R:

R₈ = R₁ + R₂ + R₃

R₈ = 4 + 10 + 6

R = 20Ω

E = I(20 + 1)

10 = 21

I = $\frac{10}{21}$

Now let's use Kirchoff's Law:

However, Vb and Vc are earthed, thus they must equal 0.

Va - 4I + 0 - 10I + 0 - 6I = Vd

Va - Vd = 20I

Va - Vd = $\frac{40}{21}$

According to Kirchhoff's voltage law, for any closed network, the voltage around a loop is equal to the total of all voltage drops in that loop and is equal to zero.

Thus, according to Kirchhoff's law, the total current flowing into and out of a junction in an electrical circuit equals the total current flowing into and out of the junction.