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In the series combination of two or more than two resistances
In series combination, current across its circuit components is always constant and in parallel combination the voltage across the circuit components is constant.
Match the Column I with Column II.
Combine three resistors 5Ω, 4.5Ω and 3Ω in such a way that the total resistance of this combination is maximum
For maximum equivalent or total resistance of the resistors must be combined in series.
R_{eq} = R_{1 }+ R_{2} = R_{3} = 5 + 4.5 + 3
= 12.5Ω
The total resistance in the parallel combination of three resistances 9Ω, 7Ω and 5Ω
In the parallel combination of three resistance's resistance is
or
= 2.02Ω
Equivalent resistance (in ohm) of the given network is
Between points A and B all resistances are combined in series
∴ R_{eq} = 3Ω + 4Ω + 5Ω + 6Ω
= 18Ω
Between points A and B all resistances are combined in series
∴ R_{eq} = 3Ω + 4Ω + 5Ω + 6Ω
= 18Ω
Which arrangement of 3Ω resistors will give a total resistance of 7Ω?
Three resistors of 3Ω each in parallel will give
R_{eq }= 3 + 1 + 3 = 7Ω.
The equivalent resistance of series combination of four equal resistors is S. If they are joined in parallel, the total resistance is P. The relation between S and P is given by S = nP. Then the minimum possible value of n is
Let Resistance of each resistor is = R
In series connection:
R_{eq1} = 4R = S
R = S/4 → (1)
In parallel connection
R_{eq2} = R4 = P
R = 4/P → (2)
Equating (1) and (2)
S/4 = 4P
S = 16P
n = 16
Five equal resistances of 10Ω are connected between A and B as shown in figure. The resultant resistance is
According to the given circuit 10Ω and 10Ω resistances are connected in series.
∴ R′ = 10 + 10 = 20Ω
Again 10Ω and 10Ω resistances are connected in series
∴ R′′ = 10 + 10 = 20Ω
R′, R′′ and 10Ω all connected in parallel than
∴
=
R_{eq }= 5Ω
The correct combination of three resistances 1Ω, 2Ω and 3Ω to get equivalent resistance 11/5Ω is
According to option (d) 2Ωand3Ω are combined in a parallel combination. Hence equivalent resistance
R' combined in series to 1Ω
∴ R_{eq} =
Hence, the combination scheme in option (d) is correct.
Equivalent resistance of the given network between points A and B is
In each segment of the combination 3Ω and 2Ω resistance are connected in series separately.
∴ R' = +3 = 6Ω and R'' = 2 + 2 = 4Ω
R' and R'' are connected in parallel
∴ For first segment
Similarly for second and third segment
Now segment is connected in series then the total resistance of combination is
R_{eq} = Re_{q1 }+ R_{eq2} + R_{eq3}
=
n resistors each of resistance R first combine to give maximum effective resistgance and then combine to give minimum. The ratio of the maximum resistance is
To get maximum equivalent resistance all resistances must be connected in series
∴ (R_{eq})_{max} = R + R + R...ntimes = nR
To get minimum equivalent resistance all resistances myst be connected in parallel.
∴
The equivalent resistance between A and B for . the circuit shown in this figure is
For a equivalent resistance betweeen A and B.
5Ω and 8Ω resistance are connected in series. R' their equivalent resistance is parallel to 6Ω
∴ R = 5 + 8 = 13Ω and
R" = 78/19
Now 4Ω, R" an d 5Ω resistances are connected in series equivalent resistance between ,A and B
∴
A copper cylindrical tube has inner radius a and outer radius b. The resistivity is ρ. The resistance of the cylinder between the two ends is
If one had considered a solid cylinder of radius b, one can suppose that it is made of two concentric cylinders of radius a and the outer part, joined along the length concentrically one inside the other.
If Ia and Ix are the currents flowing through the inner and outer cylinders
∵ I_{total} = I_{b} = I_{a} + I_{x}
⇒ VR_{b} = V/R_{a} + V/R_{x}
where R_{b} is the total resistance and R_{x} is the resistance of the tubular part.
∴
But
∴
∴
A wire of resistance 12 ohms per meter is bent to form a complete circle of radius 10cm. The resistance between its two diametrically opposite points, A and B as shown in the figure is
Wire of length 2π x 0.1 m of 12Ωm^{1} is bent to a circle.
Resistance of each part = 12 x π x 0.1
= 1.2π Ω
Total resistance = 0.6π Ω
A and B are two points on a uniform ring of resistance 15Ω. The ∠AOB = 45^{∘}. The equivalent resistance between A and B is
Resistance per unit length of ring,
Length of sections ADB and ACB are rθ and r(2π − θ)
∴ Resistance of section ADB,
and resistance of section ACB,
R_{2 }= ρr(2π − θ)
=
Now, R and R2 are connected in parallel between A and B then
=
Putting θ = 45^{∘ }= π/4 rad and R = 15Ω
=
= 1.64Ω
Two metal wires of identical dimensions are connected in series. If σ_{1} and σ_{2} are the conductivities of the metal wires respectively, the effective conductivity of the combination is
Resistance of a wire in terms of connectivity (σ) is given by
where l is the length and A is area of cross section of wire respectively.
∴ Rs = R1 + R2
⇒
where σ_{s} is the effective conductivity
σ_{s = }2σ_{1}σ_{2 }/ σ_{1 } + σ_{2}
Three resistors 2 Ω, 4 Ω and 5 Ω are combined in parallel. This combination is connected to a battery of emf 20 V and negligible internal resistance, the total current drawn from the battery is
Potential of 20V20V will be same across each resistance current
∴ Total current drawn from circuit
I = I_{1} + I_{2} + I_{3}
= 10 + 5 + 4 = 19A
Three resistors of resistances 3Ω,4Ω and 5Ω are combined in parallel. This combination is connected to a battery of emf 12V and negligible internal resistance, current through each resistor in ampere is?
Since the voltage across the circuit is constant Then current through 3Ω resistor
The current through 4 £2 resistor
and the current through 5 £2 resistor,
The reading of ammeter shown in figure is
The equivalent circuit of the given circuit will be reduced to as shown in figure.
Total resistance of the circuit
=
=
Total current in the circuit =
Reading of ammeter =
= 2.18A
Three resistances 2Ω, 4Ω, 5Ω are combined in series and this combination is connected to a battery of 12 V emf and negligible internal resistance. The potential drop across these resistances are
Let current in the circuit is I. Then total resistance in the circuit.
R = R_{1 }+ R_{2} + R_{3} = 2 + 4 + 5 = 11Ω
∴
The potential drop across 2Ω resistance
The potential drop across 4Ω resistance
The potential drop across 5Ω resistance
Hence(V_{1}, V_{2}, V_{3}) = (2.18, 4.36, 5.45)V
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