In a meter bridge circuit, the two resistances in the gap are `5 Omega` and `10 Omega`. The wire resistance is `4 Omega`. The emf of the cell connected at the ends of the wire is `5 V` and its internal resistance is `1 Omega`. What current will flow through the galvenometer of resistance `30 Omega` if the contact is made at the midpoint of wire ?

Answer

`(25)/(1357) A`

Explanation

The distribution of currents is shows in Fig. `S6.12`. Applying kirchhoff's law to closed mesh (1),(2), and (3), we have

`5(I - i_(1)) + 30 i_(g) - 2i_(1) = 0` or `5i - 7i_(1) + 30i_(g) = 0` (i)
`10(i - i_(1) - i_(g)) - 2(i_(1) + i_(g)) - 30i_(g) = 0`
or `10i - 12i_(1) - 42i_(g) = 0` (ii)
`2i_(1) + 2(i_(1) + i_(g)) + i = 5` or `4i_(1) + 2i_(g) + i = 5` (iii)
From Eqs. (i) and (ii), `i_(1) = 51i_(g)` (iv)
From Eqs. (ii) and (iii), `13i = 36i_(g) + 15` (v)
Substituting Eqs.(iv) and (v) in Eq. (iii), we get
`i_(g) = (25)/(1357) A`