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Aneli [31]
3 years ago
10

The battery has a voltage of 12V. R1 = R2 = 100Ω.

Physics
1 answer:
Margaret [11]3 years ago
4 0

Answer:

<h3> FOR PARALLEL CONNECTION</h3><h3>I1 = 0.12A</h3><h3>I2 = 0.12A </h3><h3>IT =0.24A</h3><h3>FOR SERIES CONNECTION</h3><h3>I1 = I2 = 0.06A</h3><h3>IT =0.06A</h3><h3 />

Explanation:

According to ohms law, V =ITRt

V is the supply voltage

IT is the total current flowing in the circuit

Rt is the total equivalent resistance

Given R1= R2= 100Ω

V= 12V

FOR PARALLEL CONNECTION;

To calculate the total current IT in the battery, we need to calculate the total equivalent resistance RT first. For a parallel connected circuit, the equivalent resistance in the circuit is the sum of the reciprocal of its individual resistances as shown;

\frac{1}{RT} = \frac{1}{100} + \frac{1}{100}\\\frac{1}{RT} = \frac{2}{100} \\\frac{RT}{1} = \frac{100}{2}\\

RT = 50Ω

from the equation above;

IT = V/RT

IT = 12/50

IT = 0.24A

Note that in a parallel connected circuit, different current flows through the resistances but the same voltage is across them.

IT = I1+I2

For current in resistance R1;

I1 = V/R1

I1 = 12/100

I1 = 0.12A

Since both resitance are the same, they will share the total current equally. Therefore I2 = 0.12A

FOR SERIES CONNECTION;

The total equivalent resistance in the circuit will be the sum of their individual resistances.

RT = R1+ R2

RT = 100Ω+100Ω

RT = 200Ω

IT = V/RT

IT = 12/200

IT = 0.06A

Since the resistances are connected in series, the same current will flow through them but different voltages. The total current flowing in the circuit will be the same current flowing through the resistors.

Therefore I1 = I2 = 0.06A

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  • The equivalent capacitance between point a and b is 5.87μf.
  • The charge at 20μf is 93.92 μC.
  • The charge at 6μf is 67.8 μC.
  • The charge at C and 3μf is 26.12 μC.

<h3>Sum of capacitance of C and 3μF</h3>

The sum of the capacitance is calculated as follows;

1/Ct = 1/C + 1/3

1/Ct = 1/10μf + 1/3μf

1/Ct = (3μf + 10μf)/30μf²

1/Ct = 13μf/30μf²

Ct = 30μf²/13μf

Ct = 2.31μf

<h3>Total capacitance in parallel arrangement</h3>

The total capacitance in parallel arrangement is calculated as follows;

Ct = 2.31μf + 6μf = 8.31μf

<h3>Equivalent capacitance between point a and b</h3>

1/Ct = 1/8.31μf + 1/20μf

1/Ct = 0.1703

Ct = 5.87μf

<h3>Charge flowing in each capacitor</h3>

Maximum voltage is delivered in 20μf, q = CV

<u>charge for 20μf:</u>

q = (5.87 x 16)μC

q = 93.92 μC

<h3>Equivalent capacitance for C, 3μf and 6μf</h3>

Ct = 2.31μf + 6μf = 8.31uf

<u>charge for 6μf:</u>

q = (6/8.31) x 93.92μC

q = 67.8μC

<h3>Total charge for C and 3μf</h3>

q = 93.92μC - 67.8μC = 26.12 μC

charge for C = charge 3μf = 26.12 μC

Learn more about capacitor here: brainly.com/question/14883923

#SPJ1

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