Practical 1

Greetings dear steemians, today i will love to be teaching you practical, although I have so many on my mind but i will love to take it one by one, and the one i am going to take is the wheatstone bridge.

Wheatstone bridge consist of four resistance shown below the page;

G here is the galvanometer connected across the junction A and B of the circuit. Now galvanometer here can be defined as the instrument used for measuring current and can be converted to voltmeter. E here is the dry cell of the e.m.f which drives the current through the various arms of the bridge. Now k¹ and k² are connected to the arms of the arms of the cell and galvanometer.

Now let supposed that R¹ is an unknown resistance of R² R³ and R⁴ are resistance boxes. For random values of the resistance a current will flowers through the circuit when when the keys are closed, and the galvanometers will show a deflection. By varying one or more of the resistance, a point is reached where the galvanometer show no deflection.

At this balance point, the current I¹ which flows through R¹ should also flow through R² , likewise the current I² flows through R³ and R⁴. Now since B and D are connected at the same potential and are connected across, the results will be;

P.d across AB= P.d across AD
P.d across BC = P.d across CD

Dividing both side we have:
I¹R²÷I¹R²=I¹R³÷I²R⁴

Hope you understand what I just explain thank you.........