Question

An electric lamp of resistance \[20\;\Omega \] and a resistor of resistance \[4\;\Omega \] are connected to a 6 V battery as shown in the circuit diagram.

Calculate:
 A. The total resistance of the circuit.
B. The current through the circuit.
C. The potential difference across the
1. electric lamp
2. Resistor
D. Power of the lamp.

Answer 1

Hint: The above solution can be resolved with the help of various mathematical relations of electrical circuits like , total resistance, which is the sum of resistances of individual components, voltage supply, that is the calculated by taking the fraction of current flow across the component and the magnitude of resistance of the component.


Complete step by step solution
Given:
The resistance of the electric lamp is, \[{R_1} = 20\;\Omega \].
The resistance of the resistor is, \[{R_2} = 4\;\Omega \].
The voltage supply is, V = 6 V.

A. The total resistance of the circuit is,
\[{R_T} = {R_1} + {R_2}\]
Substituting the values as,
 \[\begin{array}{l}
{R_T} = 20\;\Omega + 4\;\Omega \\
{R_T} = 24\;\Omega
\end{array}\]
 Therefore, the magnitude of total resistance is \[24\;\Omega \].

B. The current through the circuit is,
\[{I_C} = \dfrac{V}{{{R_T}}}\]
Substituting the values as,
 \[\begin{array}{l}
{I_C} = \dfrac{{6\;{\rm{V}}}}{{24\;\Omega }}\\
{I_C} = 0.25\;{\rm{A}}
\end{array}\]
Therefore, the magnitude of current through the circuit is 0.25 A.

C. 1. The Potential across the electric lamp is,
\[{V_1} = {I_C} \times {R_1}\]
Substituting the values as,
\[\begin{array}{l}
{V_1} = 0.25\;{\rm{A}} \times 20\;\Omega \\
{V_1} = 5\;{\rm{V}}
\end{array}\]
Therefore, the magnitude voltage across the electric lamp is 5 V.

C.2 The Potential across the conductor is,
\[{V_2} = {I_C} \times {R_2}\]
Substituting the values as,
\[\begin{array}{l}
{V_2} = 0.25\;{\rm{A}} \times 4\;\Omega \\
{V_2} = 1\;{\rm{V}}
\end{array}\]
Therefore, the magnitude voltage across the conductor is 1 V.

D. The power of the lamp can be calculated as,
\[P = {V_1} \times {I_C}\]
Substituting the values as,
\[\begin{array}{l}
P = 5\;{\rm{V}} \times {\rm{0}}{\rm{.25}}\;{\rm{A}}\\
P = 1.25\;{\rm{W}}
\end{array}\]
Therefore, the magnitude power through the electric lamp is \[1.25\;{\rm{W}}\].


Note: To resolve the given condition, one must remember the key formulas of electrical fundamentals like power supply, voltage supply, equivalent resistances and many more. The appropriate values of the current and resistances are to be used for a specific component to achieve the desired result.