The total resistance in the circuit is given as:
\[ R_{\text{eq}} = 4 \, \Omega. \]
Using Ohm’s law:
\[ i = \frac{V}{R_{\text{eq}}}. \]
Substitute the given values (\( V = 8 \, \text{V} \), \( R_{\text{eq}} = 4 \, \Omega \)):
\[ i = \frac{8}{4} = 2 \, \text{A}. \]
The current \( i_1 \) is determined using the current division rule:
\[ i_1 = i \cdot \frac{R_2}{R_1 + R_2}, \]
where \( R_1 = 6 \, \Omega \) and \( R_2 = 3 \, \Omega \). Substituting the values:
\[ i_1 = 2 \cdot \frac{3}{3 + 6} = 2 \cdot \frac{3}{9} = \frac{6}{9} = \frac{2}{3} \, \text{A}. \]
The current \( i_2 \) is given by:
\[ i_2 = \frac{i_1}{2}. \]
Substitute \( i_1 = \frac{2}{3} \):
\[ i_2 = \frac{\frac{2}{3}}{2} = \frac{2}{3} \cdot \frac{1}{2} = \frac{1}{3} \, \text{A}. \]
The calculated currents are:
The graph between variation of resistance of a wire as a function of its diameter keeping other parameters like length and temperature constant is
A wire of resistance $ R $ is bent into a triangular pyramid as shown in the figure, with each segment having the same length. The resistance between points $ A $ and $ B $ is $ \frac{R}{n} $. The value of $ n $ is:
Match List-I with List-II: List-I
The dimension of $ \sqrt{\frac{\mu_0}{\epsilon_0}} $ is equal to that of: (Where $ \mu_0 $ is the vacuum permeability and $ \epsilon_0 $ is the vacuum permittivity)