Advertisements
Advertisements
प्रश्न
Three cells, each of emf E but internal resistances 2r, 3r and 6r are connected in parallel across a resistor R.
Obtain expressions for (i) current flowing in the circuit, and (ii) the terminal potential differences across the equivalent cell.
Advertisements
उत्तर
`R_{AB}` = ?
`1/(R_{AB}) = 1/(2r) + 1/(3r) + 1/(6r)`
`1/R_{AB} = 1/r[(3 + 2 + 1)/6]`
`1/R_{AB} = 1/r`
`R_{AB} = r` ⇒ equivalent internal resistant
Equivalent emf of cells in parallel,
`E_{eq} = (E_1/r_1 + E_2/r_2 + E_3/r_3)/(1/r_1 + 1/r_2 + 1/r_3)`
= `(E/(2r) + E/(3r) + E/(6r))/(1/(2r) + 1/(3r) + 1/(6r))`
`E_{eq} = E[(1/(2r) + 1/(3r) + 1/(6r))/(1/(2r) + 1/(3r) + 1/(6r))]`
Eeq = E
Current flowing in the circuit,
I = `E_{eq}/(R + R_{AB})`
I = `E/(R + r)`
Potential drop across E,
V = IReq
V = `(rE)/(R + r)`
APPEARS IN
संबंधित प्रश्न
The earth’s surface has a negative surface charge density of 10−9 C m−2. The potential difference of 400 kV between the top of the atmosphere and the surface results (due to the low conductivity of the lower atmosphere) in a current of only 1800 A over the entire globe. If there were no mechanism of sustaining atmospheric electric field, how much time (roughly) would be required to neutralise the earth’s surface? (This never happens in practice because there is a mechanism to replenish electric charges, namely the continual thunderstorms and lightning in different parts of the globe). (Radius of earth = 6.37 × 106 m.)
A long straight current carrying wire passes normally through the centre of circular loop. If the current through the wire increases, will there be an induced emf in the loop? Justify.
Can the potential difference across a battery be greater than its emf?
Two non-ideal batteries are connected in parallel. Consider the following statements:-
(A) The equivalent emf is smaller than either of the two emfs.
(B) The equivalent internal resistance is smaller than either of the two internal resistances.
The following figure shows an arrangement to measure the emf ε and internal resistance r of a battery. The voltmeter has a very high resistance and the ammeter also has some resistance. The voltmeter reads 1.52 V when the switch S is open. When the switch is closed, the voltmeter reading drops to 1.45 V and the ammeter reads 1.0 A. Find the emf and the internal resistance of the battery.
A battery of emf 100 V and a resistor of resistance 10 kΩ are joined in series. This system is used as a source to supply current to an external resistance R. If R is not greater than 100 Ω, the current through it is constant up to two significant digits.
Find its value. This is the basic principle of a constant-current source.
Find the emf of the battery shown in the figure:
Two cells of emfs approximately 5 V and 10 V are to be accurately compared using a potentiometer of length 400 cm.
The internal resistance of a cell is the resistance of ______
A battery of EMF 10V sets up a current of 1A when connected across a resistor of 8Ω. If the resistor is shunted by another 8Ω resistor, what would be the current in the circuit? (in A)
