Question

You are measuring the current in a circuit that is operated on a 10 V battery. The ammeter reads 50 mA. Later, you notice that the current has dropped to 30 mA. Eliminating the possibility of a resistance change, you must conclude that the voltage has changed

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Answer 1

Given data : You are measuring the current in a circuit that is operated on a 10 V battery. The ammeter reads 50 mA. Later, you notice that the current has dropped to 30 mA.

To find : Eliminating the possibility of a resistance change, you must conclude that the voltage has changed ?

Solution : A/C to question,

• Voltage, V = 10 v
• Current, I = 50 mA

and

Let, the voltage V' be x

• Voltage, V' = x
• Current, I' = 30 mA

Now, we use ohm's law;

Definition : Ohm's law states that the current through a conductor between two points is directly proportional to the voltage across the two points.

➜ Voltage = current * resistance

➜ V = I * R

➜ R = V/I ----{1}

Similarly,

➜ R = V'/I' ----{2}

{due to eliminating the possibility of a resistance change}

From {1} and {2}

➜ V/I = V'/I'

➜ 50/10 = x/30

➜ x = (50 * 30)/10

➜ x = 1500/10

➜ x = 150 v

∴ Voltage V' is equal to 150 v.

Answer : Hence, the voltage change 150 v.

{More info :

• Voltage : Voltage is the difference in charge between two points.
• Current : Current is the rate at which charge is flowing.
• Resistance : Resistance is a material's tendency to resist the flow of charge (current).}

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Answer 2

Answer:Hence, the voltage change 150 v

Explanation:

given :

V1 = 10v

I1 = 50 ma

V2 = ?

I2 = 30ma

according to ohms law :Ohm's law states that the current through a conductor between two points is directly proportional to the voltage across the two points.

➜ Voltage = current * resistance

➜ V = I * R

v = IR

so

R = v/I

R = 10/0.05 = 200 ohm

as reistance remains the same (200)

current reduces to 30 ma

new voltage becomes:

V= IR

v = 0.03 * 200

v = 6V

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