A battery applies 45 volts to a circuit, while an ammeter reads 10 ma. later, the ammeter reading drops to 7 ma. assuming the resistance has not changed, the voltage must have

Ohm's law relates the voltage, current and resistance using the following relation:

V = I*R

where:

V is the applied voltage

I is the current

R is the resistance

From this relation, we can note that the voltage and current are directly related to each other assuming constant resistance.

This means that when the current decreases, the voltage must have decreased as well.

Now, if we want to calculate the value of the new voltage, we can follow the following steps:

We are first given that:

Voltage = 45 volts

Current = 10 ma = 10*10^-3 ampere

Substituting in the above relation:

V = IR

R = V/I = 45/10*10^-3 4500 Ω

Later, the current dropped to 7*10^-3 ampere and the resistance is kept the same.

We will again substitute in the above equation to get the value of the new voltage as follows:

V = IR = 7*10^-3 * 4500 = 31.5 volts

Based on the calculations, our conclusion that the voltage will decrease as the current decreases is verified.

This can be computed using Ohm's law relates the voltage, current and resistance using the following relation:

V = I * R

Where:

V is the applied voltage

I is the current

R is the resistance

We can say that the voltage and current are straight related to each other supposing there is continuous resistance. Meaning that when the current decreases, the voltage must have decreased as well.

So let us calculate the new voltage:

Voltage = 45 volts

Current = 10 ma = 10*10^-3 ampere

Substituting in the above relation:

V = IR

R = V/I = 45/10*10^-3 4500 Ω

Well along, the current released to 7*10^-3 ampere and the resistance is reserved the same.

So using the old equation: V = IR = 7*10^-3 * 4500 = 31.5 volts

Therefore, the voltage decreases as the current decreases.