Question

Five equal resistances each of resistance $R$ are connected as shown in the figure. A battery of $V$ volts is connected between $A$ and $B$. The current flowing in $AFCEB$ will be

• A. $\frac{3V}{R}$
• B. $\frac{V}{R}$
• C. $\frac{V}{2R}$
• D. $\frac{2V}{R}$

Answer 1

Answer: (C)

$\frac{V}{2R}$

Answer 2

By rearranging the given figure, we obtained an equivalent circuit shown in the figure below

Since the ratio of the resistance of the arms FC to CE is equal to the ratio of the resistance of arms FD to DE, therefore the given figure represents a Wheatstone Bridge.

According to the principle of Wheatstone Bridge, if FC/CE = FD/DE then no current will flow through the arm CD.

Therefore, the equivalent circuit can be given by

Now, the equivalent resistance of AFCEB, R1 = R + R = 2R

Also, the equivalent resistance of AFDEB, R2 = R + R = 2R

The equivalent resistance of the circuit, 1/Req = 1/R1 +1/R2 = 1/2R + 1/2R = 1/R

⇒ Req = R

From Ohm's law, the total current flowing through the branch AFCEB, I = Total voltage/Total resistance

⇒ I = V/2R

Hence, when five equal resistances each of resistance R are connected as shown in the figure. A battery of V volts is connected between a and b. the current flowing in AFCEB will be V/2R.

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