Calculating Total Current for Parallel Resistors
The problem asks for the total current supplied by the voltage source when two \(100\text{-}\Omega\) resistors are connected in parallel across a \(40\text{-}\text{V}\) supply.
Step 1: Calculate Equivalent Resistance
When resistors are connected in parallel, the reciprocal of the equivalent resistance (\(R_{eq}\)) is the sum of the reciprocals of individual resistances. For two resistors, the formula simplifies to:
\( R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} \)
Given \(R_1 = 100\text{ }\Omega\) and \(R_2 = 100\text{ }\Omega\):
\( R_{eq} = \frac{100\text{ }\Omega \times 100\text{ }\Omega}{100\text{ }\Omega + 100\text{ }\Omega} = \frac{10000\text{ }\Omega^2}{200\text{ }\Omega} = 50\text{ }\Omega \)
Step 2: Apply Ohm's Law
Ohm's Law states that the voltage (\(V\)) across a circuit (or component) is equal to the product of the current (\(I\)) flowing through it and its resistance (\(R\)): \(V = I \times R\). To find the current supplied by the source, we rearrange the formula to \(I = \frac{V}{R}\).
Using the supply voltage (\(V = 40\text{ V}\)) and the calculated equivalent resistance (\(R_{eq} = 50\text{ }\Omega\)):
\( I = \frac{40\text{ V}}{50\text{ }\Omega} = \frac{4}{5}\text{ A} = 0.8\text{ A} \)
Therefore, the current supplied by the voltage source is \(0.8\text{ A}\).
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 :
