Consider star connected load and two wattmeters connected as shown in Fig.1

Fig.1 Two wattmeter method for star connected load

For balanced load:

Considering the above figure 1 in which Two Wattmeter W1 and W2 are connected, the instantaneous current through the current coil of Wattmeter, W1 is given by the equation shown below.

$W_1=i_R$

Instantaneous potential difference across the potential coil of Wattmeter, W1 is given as

$W_1=e_{RN}-e_{BN}$

Instantaneous power measured by the Wattmeter, W1 is

$W_1=i_R(e_{RN}-e_{BN}).......(1)$

The instantaneous current through the current coil of Wattmeter, W2 is given by the equation

$W_2=i_Y$

Instantaneous potential difference across the potential coil of Wattmeter, W2 is given as

$W_2=e_{YN}-e_{BN}$

Instantaneous power measured by the Wattmeter, W2 is

$W_2=i_Y(e_{YN}-e_{BN}).....(2)$

Therefore, the Total Power Measured by the Two Wattmeters W1 and W2 will be obtained by adding the equation (1) and (2).

$W_1+W_2=i_R(e_{RN}-e_{BN})+i_Y(e_{YN}-e_{BN})\\ W_1+W_2=i_R\ e_{RN}+i_Y\ e_{YN}-e_{BN}(i_R+i_Y)\\ we \ know \ i_R+i_Y+i_B=0\\ W_1+W_2=i_R\ e_{RN}+i_Y\ e_{YN}+i_B\ e_{BN}\\ W_1+W_2= \ P$

                 where P is total three phase instantenous power.