Question

How many 176 Ω resistors (in parallel) are required to carry 5A on a 220 V line ?


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Answer 1

Answer:

4

Explanation:

Q 10: How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?

Ans: For x number of resistors of resistance 176 Ω,

Here Supply voltage, V = 220 V

Current, I = 5 A

Equivalent resistance of the combination = R, given as

From Ohm’s law,

∴four resistors of 176 Ω are required to draw the given amount of current.

Answer 2

S O L U T I O N :

$\underline{\bf{Given\::}}$

• Potential difference = 220 V
• Value of each resistor, (r) = 176 Ω
• Current required to flow = 5 A

$\underline{\bf{Explanation\::}}$

Let the total resistance needed in the circuit to generate a current of 5 A will be R .

As we know that formula of the resistance;

$\boxed{\bf{R = \dfrac{V}{I} }}$

A/q

→ R = V/I

→ R = 220/5

→ R = 44 Ω

Now, we have to use resistor in parallel sequence to generate a total Resistance, (R) = 44Ω .

Let the number of resistor required be n in parallel combination.

$\mapsto\tt{\dfrac{1}{R} = n\bigg(\dfrac{1}{r} \bigg)}$

$\mapsto\tt{\dfrac{1}{R} = \dfrac{n}{r}}$

$\mapsto\tt{\dfrac{1}{44} = \dfrac{n}{176}}$

$\mapsto\tt{44n = 176 }$

$\mapsto\tt{n= \cancel{176/44}}$

$\mapsto\bf{n=4}$

Thus,

The 4 resistor of 176 in parallel combination can produce the required current .