Let `V` and `I` respresent, respectively, the readings of the voltmeter and ammetre shows in Fig. `6.34`, and let `R_(V)` and `R_(V)` be their equivalent resistances. Because of the resistances of the meters, the resistnce `R` is not simply equal to `V//I`.
(a) When the circuit is connected as shows in Fig. `6.34` (a), shows that `R = (V)/(I) - R_(A)`
Explain why the true resistance `R` is always less than `V//I`.
(b) When the connections are as shows in Fig.`6.34` (b)
Show that `R = (V)/(I - (V//R_(V)))`

Explain why the true resistance `R` is always greater than `V//I`.
(c ) Show that the power delivered to the resistor in part
(i) is `IV - I^(2) R_(A)` and that in part (ii) is `IV - (V^(2)//R_(V))`

Explanation

`V = IR + IR_(A)` or `R = (V)/(I) - R_(A)`
The true resistance `R` is always less then the reading `(V//I)` because in the circuit, the ammeter's resistance causes the current to be less than the actual. Thus, the smaller current required resistance `R` to be calculated larger than what it should be.
(b) `I = (V)/(R ) + (V)/(R_(V))` or `R = (VR_(V))/(IR_(V) - V) = (V)/(I - V//R_(V))`
Now the current measured is greater than that through the resistor, so `R = V//I_(R)` is always greater than `V//I`.
(c ) (i) `P = I^(2)R = I^(2)(V//I - R_(A)) = IV - I^(2)R_(A)`
(ii) `P = V^(2)//R = V (I- V//R_(V)) = IV - V^(2)//R_(V)`