Parallel Combination of Cells
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Derive an expression for three resistances connected in parallel.
- V3r
- V4r
- V2r
- V6r
- 10 Ω
- 1 Ω
- 5 Ω
- 25 Ω
- 4 Ω
- 2 Ω
- 8 Ω
- 6 Ω
- 4Ω
- 4.8Ω
- 2Ω
- 1.2Ω
Draw a schematic diagram of a circuit consisting of a battery of of each, a combination of three resistors of , and connected in parallel, a plug key, and an ammeter, all connected in series. Use this circuit to find the value of the current through each resistor.
Output voltage across load is (approximately)
- 6.2 V
- 3.2 V
- 5.2 V
- 4.9 V
- 3V
- 3.75V
- 3.25V
- 3.5V
- 2
- 1
- 0.8
- 0.4
- 1 Ω
- 23 Ω
- 25 Ω
- 52 Ω
- 14πϵ0[Q1r+Q2r+Q3r]
- 14πϵ0[Q1+Q2r+Q3c]
- 14πϵ0[Q1a+Q2b+Q3c]
- 14πϵ0c×[Q1+Q2+Q3]
- Only a capacitor
- Resistor and capacitor
- Only resistor
- Resistor and an inductor
- R
- R(√3−1)
- 3R
- R(√3+1)
A single battery fo emf E and internal resistance r is equivalent to a parallel combination of two batteries of emfs E1=2V and E2=1.5V and internal resistances r1=1Ω and r2=2Ω respectively, with polarities as shown in the figure.
(IIT-JEE 1997)
E=56V
r = 1.5 Ω
E = 1.2 V
r=23Ω
- 1 A
- 2 A
- 3 A
- 4 A
In the figure shown, the current in the 10 V battery is close to :
- 0.71 A from positive to negative terminal
- 0.42 A from positive to negative terminal
- 0.21 A from positive to negative terminal
- 0.36 A from negative to positive terminal
Six identical cells each of and internal resistance are connected in parallel to form a battery. An external resistance of is connected to this battery. Draw the circuit diagram of the above arrangement. What is the current in resistor.
- 0.44V
- 1.55V
- 2.66V
- 1.33V
- 2A
- 3A
- 3.5A
- 2.5A
- 23ϵ0AEd
- ϵ0AEd
- 4ϵ0AEd
- 34ϵ0AEd
A potential difference has been applied across A and B. On closing the switch S, what will be the effect on the readings of voltmeters ?
- V1 increases
- V2 decreases
- V2 and V3 both increase
- One of V2 and V3 increases and the other decreases
- 0A
- 10A
- 0.1A
- 1A
- R=25 Ω, L=2.5 H, C=45 μ F
- R=15 Ω, L=3.5 H, C=30 μ F
- R=25 Ω , L=1.5 H, C=45 μ F
- R=20 Ω , L=1.5 H, C=35 μ F
- (2n+1)C
- n2C
- (n−1)n2C
- (n+1)n2C
- (800 ± 1) Ω
- (800 ± 7) Ω
- (200 ± 7) Ω
- (200 ± 1) Ω
Currently the switch is in position A and steady current is flowing through circuit. Capacitor is uncharged. At t=0, switch is instantly moved to position B.
Find the current in the LC circuit.(Given that ω = 1√LC)
- VRsin(ωt)
- VRcos(ωt)
- VRcos(ωt+π5)
- VRcos(ωt+π8)