4 cells each of emf 2 V and internal resistance of `1Omega` are connected in parallel to a load resistor of `2Omega`. Then the current through the load resistor is

Options

A. `2 A`
B. `1.5 A`
C. `1 A`
D. `0.888 A`

Explanation

To find the current through the load resistor when 4 cells (each with an EMF of 2 V and internal resistance of 1 Ω) are connected in parallel to a load resistor of 2 Ω, we can follow these steps:

### Step 1: Calculate the total EMF and internal resistance of the cells in parallel

When cells are connected in parallel, the EMF remains the same as that of a single cell, while the total internal resistance (R_internal) can be calculated using the formula for resistors in parallel:

\[ \frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4} \]

For our case, since all internal resistances are equal (1 Ω):

\[ \frac{1}{R_{total}} = \frac{1}{1} + \frac{1}{1} + \frac{1}{1} + \frac{1}{1} = 4 \]

Thus,

\[ R_{total} = \frac{1}{4} \, \Omega = 0.25 \, \Omega \]

### Step 2: Determine the equivalent circuit

Now we have:
- EMF (E) = 2 V
- Total internal resistance (R_internal) = 0.25 Ω
- Load resistance (R_load) = 2 Ω

### Step 3: Calculate the total resistance in the circuit

The total resistance (R_total) in the circuit is the sum of the internal resistance of the cells and the load resistance:

\[ R_{total} = R_{internal} + R_{load} = 0.25 \, \Omega + 2 \, \Omega = 2.25 \, \Omega \]

### Step 4: Calculate the total current in the circuit using Ohm's Law

Using Ohm's Law, the total current (I_total) flowing from the cells can be calculated as:

\[ I_{total} = \frac{E}{R_{total}} = \frac{2 \, V}{2.25 \, \Omega} = \frac{8}{9} \, A \approx 0.89 \, A \]

### Step 5: Calculate the current through the load resistor

Since the load resistor is in parallel with the internal resistances of the cells, the voltage across the load resistor is the same as the voltage across the internal resistances. The current through the load resistor (I_load) can be calculated using Ohm's Law:

\[ I_{load} = \frac{V_{load}}{R_{load}} = \frac{E - I_{total} \cdot R_{internal}}{R_{load}} \]

First, we need to find the voltage across the load resistor:

\[ V_{load} = E - I_{total} \cdot R_{internal} = 2 \, V - \left(\frac{8}{9} \, A \cdot 0.25 \, \Omega\right) = 2 \, V - \frac{2}{9} \, V = \frac{16}{9} \, V \]

Now, substituting back to find the current through the load resistor:

\[ I_{load} = \frac{V_{load}}{R_{load}} = \frac{\frac{16}{9} \, V}{2 \, \Omega} = \frac{16}{18} \, A = \frac{8}{9} \, A \approx 0.89 \, A \]

### Final Answer

The current through the load resistor is approximately \( 0.88 \, A \).

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