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sineoko [7]
2 years ago
11

A circuit supplied with 110V carries 5 Amps current. Calculate the Resistance value of the circuit?

Physics
1 answer:
IgorC [24]2 years ago
6 0

Answer:

The value is R =  22 \  \Omega

Explanation:

From the question we are told that

     The voltage is  V  =  110 \  V

    The current is  I  =  5 \  A

Generally the resistance value is mathematically represented as

         R =  \frac{V}{I}

=>      R =  \frac{110}{5}

=>      R =  22 \  \Omega

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Answer:

(a) Charge of 4.25 μF capacitor is 35.46 μC.

(b) Charge of 1.13 μF capacitor is 10.05 μC.

(c) Charge of 2.85 μF capacitor is 25.36 μC.

Explanation:

Let C₁ , C₂ and C₃ are the capacitor which are connected to the battery having voltage V. According to the problem, C₁ and C₂ are connected in parallel. There equivalent capacitance is:

C₄ = C₁ + C₂

Substitute 1.13 μF for C₁ and 2.85 μF for C₂ in the above equation.

C₄ = ( 1.13 + 2.85 ) μF = 3.98 μF

Since, C₄ and C₃ are connected in series, there equivalent capacitance is:

C₅ = \frac{C_{3}C_{4}  }{C_{3} + C_{4}  }

Substitute 4.25 μF for C₃ and 3.98 μF for C₄ in the above equation.

C₅ = \frac{4.25\times3.98 }{4.25 + 3.98  }

C₅ = 2.05 μF

The charge on the equivalent capacitance is determine by the relation :

Q = C₅ V

Substitute 2.05 μF for C₅ and 17.3 volts for V in the above equation.

Q = 2.05 μF x 17.3  = 35.46 μC

Since, the capacitors C₃ and C₄ are connected in series, so the charge on these capacitors are equal to the charge on the equivalent capacitor C₅.

Charge on the capacitor, C₃ = 35.46 μC

Charge on the capacitor, C₄ = 35.46 μC

Voltage on the capacitor C₄ = \frac{Q}{C_{4} } = \frac{35.46\times10^{-6} }{3.98\times10^{-6}} = 8.90 volts

Since, C₁ and C₂ are connected in parallel, the voltage drop on both the capacitors are same, that is equal to 8.90 volts.

Charge on the capacitor, C₁ = C₁ V = 1.13 μF x 8.90 = 10.05 μC

Charge on the capacitor, C₂ = C₂ V = 2.85 μF x 8.90 = 25.36 μC

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