Question

A D.C wheatstone bridge has a 6.0V supply connected between point A and C what will be the potential difference between point B and D when the resistance in the bridge arms are AB 10(ohms), BC(20ohms), CD(60ohms), AD(31ohms)

Answer 1

Draw the situation:

According to Kirchoff's laws we can write:

$$I=I_1+I_2,\\ I_1(10+20)=6,\\ I_2(60+31)=6.\\ I_1=0.2\text{ A},\\ I_2=0.066\text{ A},\\ I=0.266\text{ A}.\\$$

Now imagine that we put a voltmeter between B and D. Then, using the same Kirchoff's laws:

$$V_{BD}=V_D-V_B=\\ =(0.066\cdot31-0.066\cdot60)-(0.2\cdot10-0.2\cdot20)=\\ =0.086\text{ V}.$$