Question

(a) Find the potential difference between `A` and `B`

(b) Calculate the current flowing in the circuit and the potential difference across terminals of each cell.

( c) Calculate the potential difference across terminals of each cell.

(d) Calculate the potential difference between `A` and `B`

( e) Calculate the current in each resistor and terminal potential difference across each cell.

(f) Calculate potential difference across each cell.

(g) A battery of `6` cells each of emf `2 V` and internal resistance `0.5 Omega` is being changed by `DC` mains of emf `220 V` by using an external resistance of `10 Omega`. Calculate.
(i) the charging current
(ii) the potential difference across the battery.
.

Answer 1

(a) `i = (10)/(2 + 8) = 1 A`
`V_A - V_B = i xx 8 = 1 xx 8 = 8 V`
OR
`V_A - V_B = E - ir`
=`10 - 1 xx 2 = 8 V`
(Cell is supplying current)
.
(b) `i = (10 + 10)/(2 +1 + 7) = 2 A`
Cell `X` : `V_A - V_B = 10 - 2 xx 2 = 6 V`
Cell `Y` : `V_B - V_C = 10 - 2 xx 1 = 8 V`
(Cell are supplying current)
.
( c) Net emf `= 18 - 12 = 6 V`
Net resistance `= 3 +2 + 1 = 6 Omega`
`i = (6)/(6) = 1 A`
Cell `X` : It is taking current
`V_A - V_B = 12 + 1 xx 3 = 15 V`
Cell `Y` : It is supplying current
`V_C - V_B = 18 - 1 xx 2 = 16 V`.
.
(d) These cells are in series because same current flows in them, cells are opposing.
`i = (30 - 10)/(5 + 15) = 1 A`
Cell `X` : (supplying current), `V_A - V_B = 30 - 1 xx 5 = 25 V`
OR
Cell `Y` : (taking current), `V_A - V_B = 10 + 1 xx 15 = 25 V`.
.
( e) Net emf `= 20 - 10 = 10 V`
Net resistance `= 2 + 2 + 1 + 5 = 10 Omega`
`i = (10)/(10) = 1 A`
`p.d`. across `AB = 1 xx 2 = 2 V`
Cell `X` : (supplying current), `V_A - V_D = 20 - 1 xx 2 = 18 V`
Cell `Y` : (taking current), `V_B - V_C = 10 + 1 xx 1 = 11 V`
`i_1 = (2)/(4) = (1)/(2) A`
`i_2 = (2)/(6) = (1)/(3) A`
`i_3= (2)/(12) = (1)/(6) A`

(f) Net emf `= 8 - 4 = 4 V`
Net resistance `= 0.5 + 1 + 2 + 4.5 = 8 Omega`
`i = (4)/(8) = (1)/(2) A`
Cell `X` : (supplying current), `V_B - V_A = 8 - (1)/(2) xx 1 = 7.5 V`
Cell `Y` : (taking current), `V_B - V_C = 4 + (1)/(2) xx 0.5 = 4.25 V`
.
(g) Net emf of battery `= 6 xx 2 = 12 V`
Net resistance of battery `= 6 xx 0.5 = 3 Omega`
(i) `i = (220 - 12)/(3 + 10) = (208)/(13) = 16 A`
(ii) Battery : (taking current)
`V_A - V_B = 12 + 16 xx 3 = 60 V`.
.