Question

Two resistances of 2 ohm each are connected in parallel across a battery of 5 volt, the current in the circuit is A) 0.5 A B) 5 A C)2.5A D)0.8A
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Answer 1

Answer:

Answer: 5A

Explanation:

Given:

V= 5v ;  R1= 2 Ω ; R3= 2 Ω

The given question can be solved by using two methods:-

Method #1: By adding the current passing through each resistor

Current passing through R1,

I1 = V/R1 = 5/2 = 2.5A

Current passing through R2,

I2 = V/R2 = 5/2 = 2.5A

Total Current, I = I1+I2 = 2.5 +2.5 = 5A

Method #2: By Calculating the Equivalent Resistance

$\frac{1}{R_{p} } = \frac{1}{R_{1} } + \frac{1}{R_{2} } \\\frac{1}{R_{p} } = \frac{1}{2} + \frac{1}{2} \\\frac{1}{R_{p} } = \frac{1}{1}$

⇒$R_{p} = 1$ Ω

Total Current =

$I = V/R_{p} \\I= 5/ 1\\I = 5A$

Answer 2

The current in the circuit is 5 A and option (B) is correct.

Explanation:

Given:

Two resistances of 2 ohm each are connected in parallel across a battery of 5 volt.

To Find:

The current in the circuit.

Solution:

As given-Two resistances of 2 ohm each are connected in parallel across a battery of 5 volt.

Let the two resistors $R_{1}$ and $R_{2}$ of 2 ohm each are connected in parallel.

$R_{1} = 2\Omega \ and\ R_{2} =2\Omega$

In parallel combination,

$\frac{1}{R_{eq} } =\frac{1}{R_{1} } +\frac{1}{R_{2} }$

$\frac{1}{R_{eq} } =\frac{1}{2 } +\frac{1}{2 }$R

$\frac{1}{R_{eq} } =\frac{2}{2 } =1$

$R_{eq} =1\Omega$

Battery Voltage V=5 Volt.

Let  the current in the circuit is I amp.

Using Ohm's law

$V=I\times R_{eq}$

$5=I\times 1$

$I=5 A$

Thus,the current in the circuit is 5 A and option (B) is correct.

PROJECT CODE#SPJ3