Question

A current of one 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 the resistance of$10\Omega$ is connected in parallel with this series combination, what change (if any) in the current flowing through $5\Omega$ conductor and potential difference across the lamp will take place? Give reason.

Answer 1

Step 1: Given data

Electric current flowing through the circuit initially, $i=1A$

Resistance of the conductor, $R_c=5\Omega$

Potential difference in the battery, $V=10v$

Resistance of the resistor connected parallel to the given circuit, $R_r=10\Omega$

Resistance of the bulb, $R_b=?$

Electric current flowing through the parallel circuit, $I_p=?$

Step 2: Assumptions

Total resistance in series$=R_s$

Total resistance in parallel$=R_p$

Step 3: Calculation of the resistance of the electric lamp

The circuit diagram of the given circuit is shown below

Since the resistor and the bulb are connected in series, therefore the total resistance in this series circuit will be equal to the sum of the resistance of the resistor and the bulb.

$R_s=R_b+R_r$ ……………………..(a)

Substituting the given value of the resistance of the resistor in equation (a), we get

$R_s=R_b+5$ ……………………….(b)

Applying ohm's law to the given circuit, we get

$V=i{R}_s$ ……………………………(c)

Using equation (b) and the given value of electric current in equation (c), we get

$10v=1A\times (R_b+5)$

$$\begin{gathered} 10=R_b+5 \\ {R}_b=5\Omega \end{gathered}$$

Hence, the resistance of the bulb is equal to $5\Omega$

Step 4: Calculation of the resistance of the parallel circuit

The circuit diagram of the given parallel combination is shown below

In the given circuit, the resistances are in parallel and therefore the equivalent resistance of this parallel circuit is given by

$$\begin{gathered} \frac{1}{R_p}=\frac{1}{R_b+5}+\frac{1}{10} \\ \frac{1}{R_p}=\frac{1}{5+5}+\frac{1}{10} \\ \frac{1}{R_p}=\frac{1}{10}+\frac{1}{10} \\ \frac{1}{R_p}=0.2 \\ {R}_p=5\Omega \end{gathered}$$

Hence, the equivalent resistance of the given parallel circuit is equal to $5\Omega$

Step 5: Calculation of the current flowing through the parallel circuit

Applying ohm's law to the given parallel circuit, we get

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

Substituting the given value of potential difference and the equivalent resistance of this parallel circuit in the above equation, we get

$$\begin{gathered} I_p=\frac{10v}{5\Omega } \\ {I}_p=2A \end{gathered}$$

Hence, the magnitude of electric current flowing through the given parallel circuit is equal to $2A$

Step 6: Depicting whether the current flowing through the $5\Omega$ conductor changes or not

In the case of a parallel circuit, the electric current is divided equally in both arms.

The total current flowing through the parallel circuit is $2A$which will be divided equally in both the arms of the parallel combination.

The electric current flowing through the conductor, resistor and the bulb will be equal to $1A$

Hence, there will be no change in the electric current flowing through the conductor of resistance $5\Omega$

Step 7: Depicting whether there will be any change in potential difference across the lamp

The potential difference ($V_l$) across the lamp is equal to the product of the electric current $\left(i_l\right)$flowing through the lamp and the resistance ($R_l$) of the lamp.

$V_l=i_l\times R_l$

The electric current flowing through the lamp and the resistance of the lamp remain the same as was before connecting a resistance parallel to it.

$i_l\&R_l\to cons\tan t\Rightarrow V_{l}iscons\tan t$

Hence, the potential difference across the lamp remains the same.