In the figure let us assume the voltmeter to be ideal (internal resistance R_{in}=infinite).

So current Iv=0A

I= I_{A}+I_{V}

I=I_{A} [I_{V}=0A]

V=V_{A}+V_{K}

V=I×R_{A}+I×R_{unknown}

Ammeter is not ideal here.

V=I(R_{A}+ R_{unknown})

V/I= R_{A}+ R_{unknown }[V/I= Measured value of resistance]

**R _{M}=R_{A}+ R_{unknown}**

So the measured value of resistance is greater than the unknown value of resistance exceed by the value of R_{A}.

**Error calculation-**

e= R_{known}-R_{m}/ R_{known}

**or **e=R_{A}/ R_{known}

If R ≅0, then Rm= R_{known }(e ≈0)

I=I_{V}+I_{K}

I=V/R_{V}+V/R_{K}

I=V(1/R_{V}+1/R_{K})

I/V=(1/R_{V}+1/R_{K})

I/RM=I/RV+I/RK

I/RM=RV+ R_{known}/RV R_{known}

RM= RV R_{known}/ RV+ R_{known}

If the voltmeter is not ideal then measured value is R_{known.}

Also Read-Wheatstone bridge

Also Read- Method of substitution