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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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Answer:

Net electric field, E_{net}=91406.24\ N/C

Explanation:

Given that,

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E_1=65917.96\ N/C

Electric field due to charge 2 is given by :

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The point charges have opposite charge. So, the net electric field is given by the sum of electric field due to both charges as :

E_{net}=E_1+E_2

E_{net}=65917.96+25488.28

E_{net}=91406.24\ N/C

So, the electric field strength at the midpoint between the two charges is 91406.24 N/C. Hence, this is the required solution.

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4 years ago
I select true helllpppp me
yanalaym [24]

Answer:

your right answer is true

hope it helps you

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