Question

An ammeter is connected in series with a resistor of unknown resistance R and a DC power supply of known emf e. A student uses the ammeter to measure the current I and calculates the power P dissipated by the resistor using I and e. The student later discovers that the ammeter is not an ideal ammeter. Which of the following correctly compares the actual power dissipated by the resistor to the value of power P calculated by the student and explains why? (A) The actual power dissipated by the resistor is greater than P because the ammeter had some resistance (B) The actual power dissipated by the resistor is less than P because the ammeter had some resistance. (C) The actual power dissipated by the resistor is greater than P because the ammeter should have been connected in parallel to the resistor. (D) The actual power dissipated by the resistor is less than P because the ammeter should have been connected in parallel to the resistor. (E) The actual power dissipated by the resistor is equal to P because the ammeter was connected properly in series.

Answer 1

B:The actual power dissipated by the resistor is less than P because the ammeter had some resistance.

Step-by-step explanation:

Here,power has been calculated using current I and total EMF \ε . So,P=EMF*current= ε I will represent total power dissipated in resistor and ammeter.

Now, this total power P has been dissipated in both resistor and ammeter. So, power dissipated in resistor must be less than P as some power is also dissipated in ammeter because it has non-zero resistance.

So, the answer is B:The actual power dissipated by the resistor is less than P because the ammeter had some resistance.

Note that option A,C and E are ruled out as they state power dissipated by resistor is greater than or equal to P which is false.

Also,option D is ruled out as ammeter is connected in series.