Question

two resistors are connected in parallel .the voltage applied is 220v. the current taken is 30A. the power dissipated in one of the resistors is 1600w. find the value of the two resistors?​

Answer 1

Answer:

the pic is not that clear

Explanation:

I am not perfectly sure about this solution but hope this is correct

Answer 2

Answer:

• The value of resistor C= $\frac{121}{4}$Ω
• The value of resistor R₂ = $\frac{242}{25}$Ω

Explanation:

Given,

• Voltage applied = 220 Volt
• Current = 30 A
• Power dissipated in one of the resistor = 1600 W

Formulae:

V = IR     ...............................(1)

Since the 2 resistors are connected in parallel

$\frac{1}{R_{eq} } = \frac{1}{R_{1} } + \frac{1}{R_{2} }$     ..................(2)

Solution:

V = I R

220 = 30 × R

R = $\frac{22}{3}$

• R is the equivalent of R₁ and R₂
• The voltage in the both the resistors are same.

To find R₁:

P = V I

$P = V(\frac{V}{R})$

$P = \frac{V^{2} }{R}$

$R_{1} = \frac{220^{2} }{1000}$

R₁ = $\frac{121}{4}$Ω

To find R₂ use equation (2)

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

$\frac{3}{22} } = \frac{4}{121 } + \frac{1}{R_{2} }$

$\frac{1}{R_{2} } = \frac{3}{22} - \frac{4}{121}$

$\frac{1}{R_{2} } = \frac{25}{242}$

$R_{2} = \frac{242}{25}$Ω

Answer: The value of resistor R₁ and R₂ is $\frac{121}{4}$ Ω and $\frac{242}{25}$ Ω.

#SPJ2