Question

A current of 1 ampere flows in a series circuit containing an electric lamp and a conductor of 5 $\Omega$ when connected to a 10 V battery. Calculate the resistance of the electric lamp. Now if a resistance of 10 $\Omega$ is connected in parallel with this series combination, what change (if any) in current flowing through 5 $\Omega$ conductor and potential difference across the lamp will take place ? Give reason. Draw circuit diagram.

Answer 1

Hint : Use Ohm's law and series and parallel combination of resistors.

Step 1 : Calculation of Total Resistance in circuit

Let the resistance of lamp be $R_1$

Resistance of conductor $2$ be $R_2$

As $R_1$ and $R_2$ are in series so total resistance will be

$R_s$ = $R_1+$$5$

Step 2: Applying Ohm's law to find $R_1$

Given that $I=$ $1A$

By Ohm's law $V=$$I\times R$

$10=I\times (R_1+5)$

$10=1\times (R_1+5)$

$R_1=5\Omega$

Step 3: Calculation of change when 10$\Omega$ is connected in parallel with series combination

When $10\Omega$ is connected in parallel with the series combination the equivalent circuit is shown in figure 2

So, $\frac{1}{R_p}=$ $\frac{1}{10}+\frac{1}{10}$

$R_p=$$5\Omega$

Step 4: Calculation of total current

Current flowing in the circuit is

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

$I=$$\frac{10}{5}=2A$

Step 5: Calculation of current through lamp and $5\Omega$

As there is symmetry in the circuit, current distributes equally in both the wires will distributes equally.

So current in lamp and $5\Omega$ resistor is $1A$.

Final Step :

There is no change in current in $5\Omega$ resistance when $10\Omega$ is connected in parallel.

As there is no change in current, so there will be no change in potential across the lamp.