Parallel Resistor Current Calculation
The problem asks for the total current supplied by the voltage source when two resistors are connected in parallel across it.
Identify Given Values
- Resistance of each resistor, \(R_1 = R_2 = 100\ \Omega\).
- Voltage supply, \(V = 40\ V\).
Calculate Equivalent Resistance
For resistors in parallel, the equivalent resistance (\(R_{eq}\)) is calculated using the formula:
\( \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \)
Substituting the values:
\( \frac{1}{R_{eq}} = \frac{1}{100\ \Omega} + \frac{1}{100\ \Omega} = \frac{2}{100\ \Omega} \)
Therefore, the equivalent resistance is:
\( R_{eq} = \frac{100\ \Omega}{2} = 50\ \Omega \)
Calculate Total Current
Using Ohm's Law, the total current (\(I\)) supplied by the voltage source is:
\( I = \frac{V}{R_{eq}} \)
Substituting the values:
\( I = \frac{40\ V}{50\ \Omega} \)
\( I = 0.8\ A \)
Conclusion
The current supplied by the voltage source is 0.8 A.
Important Questions from Combination of Resistors — Series and Parallel
Consider two resistors, $R_1$ and $R_2$, connected in series to a DC voltage source. Which of the following statements accurately describes the distribution of current and voltage across these resistors?
A cell of negligible resistance and e.m.f 2 volt is connected to series combination of 2 ohm, 3 ohm and 5 ohm. The potential difference across the 3 ohm resistance is:
Two bulbs A, of (100w, 100v), and B of (60 w, 100v) are connected in series and across the series combination 200 v is applied. Which bulb will be fused?
3 resistors of 3 ohm each connected in series. What is the mean values of resistors?
The equivalent resistance of the resistances (two) joined in parallel is 6/5 Ω. When one of the resistance wire is broken, the effective resistance becomes 2Ω. The resistance of the wire that got broken was :
