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Bess [88]
3 years ago
10

The current in a series circuit is 19.3 A. When an additional 7.40-Ω resistor is inserted in series, the current drops to 13.4 A

. What is the resistance in the original circuit?
Physics
1 answer:
Bogdan [553]3 years ago
5 0

Answer:

16.8ohms

Explanation:

According to ohm's law which states that the current passing through a metallic conductor at constant temperature is directly proportional to the potential difference across its ends.

Mathematically, V = IRt where;

V is the voltage across the circuit

I is the current

R is the effective resistance

For a series connected circuit, same current but different voltage flows through the resistors.

If the initial current in a circuit is 19.3A,

V = 19.3R... (1)

When additional resistance of 7.4-Ω is added and current drops to 13.4A, our voltage in the circuit becomes;

V = 13.4(7.4+R)... (2)

Note that the initial resistance is added to the additional resistance because they are connected in series.

Equating the two value of the voltages i.e equation 1 and 2 to get the resistance in the original circuit we will have;

19.3R = 13.4(7.4+R)

19.3R = 99.16+13.4R

19.3R-13.4R = 99.16

5.9R = 99.16

R= 99.16/5.9

R = 16.8ohms

The resistance in the original circuit will be 16.8ohms

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The work done by an external force to move a -6.70 μc charge from point a to point b is 1.20×10−3 j .
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Answer:

108.7 V

Explanation:

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where

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The variation of kinetic energy of the charge is equal to the sum of the work done by the two forces:

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In this problem, we have

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4 0
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A parallel-plate capacitor has square plates that are 8.00cm on each side and 4.20mm apart. The space between the plates is comp
Sergeu [11.5K]

Answer:

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Explanation:

Given that

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A= a ² = 64 cm²

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C_2=\dfrac{K_2\varepsilon _oA}{d_2}

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U=\dfrac{CV^2}{2}

V= 76 V

U=\dfrac{4.5\times 10^{-11}\times 76^2}{2}\ J

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