Introduction:
In electrical circuits, power is calculated by multiplying voltage and current. The accuracy of power measurement depends on the accuracy of both voltage and current measurements. In this scenario, the error in measuring power is given as 10%. We have to calculate the error in measuring the current to reduce the error in power measurement to 5%.

Calculation:
Let's assume the power measured with 10% error is P1, and the power measured with 5% error is P2. We can write the following equations:

P1 = V1 * I1 (where V1 = 10V or -0.2V, and I1 = 5A or -0.4A)
P2 = V2 * I2 (where V2 = 10V or -0.2V, and I2 = (1 + x) * I1 or (1 - x) * I1)

Here, x is the percentage error in measuring the current, and we have to find its value.

We know that the relative error in power measurement is given by the sum of relative errors in voltage and current measurements:

Error in power measurement = Error in voltage measurement + Error in current measurement

For P1, the relative error in power measurement is 10%. So, we can write:

Error in P1 = Error in V1 + Error in I1
10% = Error in V1 + Error in I1

Since the error in voltage measurement is negligible (given as 0%), we can write:

Error in I1 = 10%

Similarly, for P2, we have:

Error in P2 = Error in V2 + Error in I2
5% = Error in V2 + Error in I2

Again, the error in voltage measurement is negligible. So, we can write:

Error in I2 = 5%

Substituting the values of I1, I2, and V1, V2 in the above equations, we get:

10% = 10% + Error in I1
Error in I1 = 0%

5% = 10% + Error in I2
Error in I2 = -5%

Therefore, to reduce the error in power measurement from 10% to 5%, we have to measure the current with an error of -5%. This means that the actual value of current should be 5% less than the measured value. For example, if the measured value of current is 5A, the actual value will be 4.75A.