Question

An electrician not aware of the colour coding of resistors connected two resistors A and B in series to a 6V battery of internal resistance $3\Omega$ and an ammeter. The ammeter connected in the circuit was not working and hence he disconnected the ammeter from the circuit. The sequence of the colour bands on resistor. A is yellow, violet and brown while that on resistor B is red, violet and black respectively. By using the colour coding of resistors, help the electrician to determine the current flowing through the circuit.

• A. $12$ mA
• B. $24$ mA
• C. $48$ mA
• D. $32$ mA

Answer 1

The correct option is A $12$ mA

Given: resistors A and B connected in series to a $6V$ battery of $3\Omega$ internal resistance. The sequence of the colour bands on resistor. A is yellow, violet and brown while that on resistor B is red, violet and black respectively.

To find the current flowing through the circuit.

Solution:

According to the question the A and B are 3-band resistors, so the thirdcode will be a multiplier.

And using the standard resistor color code table the value of:

yellow is 4, violet is 7, brown is $\times {10}^1$, red is 2, black is $\times {10}^0$.

The corresponding value of A and B are:

A= $47\times {10}^1=470\Omega$, B= $27\times {10}^0=27\Omega$

Now A and B are in series are also the internal resistance will also be in series. So the effective resistance of the circuit will be $R_{eq}=470+27+3=500\Omega$.

We know according to Ohm's Law,

$V=IR⟹I=\frac{V}{R}⟹I=\frac{6}{500}=0.012A$

Or the current flowing through the circuit = $12mA$