Advertisements
Advertisements
Question
Determine the current drawn from a 12 V supply with internal resistance 0.5 Ω by the infinite network shown in the figure. Each resistor has 1 Ω resistance.

Advertisements
Solution
The resistance of each resistor connected in the given circuit, R = 1 Ω
Equivalent resistance of the given circuit = R’
The network is infinite. Hence, equivalent resistance is given by the relation,
∴ R' = `2 + "R'"/(("R'" + 1))`
`("R'")^2 - 2"R'" - 2 = 0`
R' = `(2 ± sqrt(4 + 8))/2`
= `(2 ± sqrt12)/2`
= `1 ± sqrt3`
Negative value of R’ cannot be accepted. Hence, equivalent resistance,
R' = `(1 + sqrt3)` = 1 + 1.73 = 2.73 Ω
Internal resistance of the circuit, r = 0.5 Ω
Hence, total resistance of the given circuit = 2.73 + 0.5 = 3.23 Ω
Supply voltage, V = 12 V
According to Ohm’s Law, current drawn from the source is given by the ratio, `12/3.23` = 3.72 A
RELATED QUESTIONS
Kirchhoff's voltage law and current law are respectively in accordance with the conservation of .................................. .
- charge and momentum
- charge and energy
- energy and charge
- energy and momentum
Kirchhoff's junction law is equivalent to .............................
(a) conservation of energy.
(b) conservation of charge
(c) conservation of electric potential
(d) conservation of electric flux
The current is drawn from a cell of emf E and internal resistance r connected to the network of resistors each of resistance r as shown in the figure. Obtain the expression for
- the current draw from the cell and
- the power consumed in the network.

ε1 and ε2 are two batteries having emf of 34V and 10V respectively and internal resistance of 1Ω and 2Ω respectively. They are connected as shown in the figure below. Using Kirchhoff’s Laws of electrical networks, calculate the currents I1 and I2.

Given the resistances of 1 Ω, 2 Ω, 3 Ω, how will be combine them to get an equivalent resistance of (11/3) Ω?
Given the resistances of 1 Ω, 2 Ω, 3 Ω, how will be combine them to get an equivalent resistance of (11/5) Ω?
Given the resistances of 1 Ω, 2 Ω, 3 Ω, how will be combine them to get an equivalent resistance of (6/11) Ω?
Determine the equivalent resistance of networks shown in Fig.

In the given circuit, assuming point A to be at zero potential, use Kirchhoff’s rules to determine the potential at point B.

Consider the circuit shown in the figure. Find (a) the current in the circuit (b) the potential drop across the 5 Ω resistor (c) the potential drop across the 10 Ω resistor (d) Answer the parts (a), (b) and (c) with reference to the figure.


A capacitor of capacitance 8.0 μF is connected to a battery of emf 6.0 V through a resistance of 24 Ω. Find the current in the circuit (a) just after the connections are made and (b) one time constant after the connections are made.
In the circuit shown in the figure below, E1 and E2 are two cells having emfs 2 V and 3 V respectively, and negligible internal resistance. Applying Kirchhoff’s laws of electrical networks, find the values of currents l1 and I2.

Twelve wires each having a resistance of 3 Ω are connected to form a cubical network. A battery of 10 V and negligible internal resistance is connected across the diagonally opposite corners of this network. Determine its equivalent resistance and the current along each edge of the cube.
The instrument for the accurate measurement of the e.m.f of a cell is ______.
While measuring the length of the rod by vernier callipers, the reading on the main scale is 6.4 cm and the eight divisions on vernier is in line with marking on the main scale division. If the least count of callipers is 0.01 and zero error - 0.04 cm, the length of the rod is ______.
Kirchhoff s second law is based on the law of conservation of ______
Kirchhoff’s junction rule is a reflection of ______.
- conservation of current density vector.
- conservation of charge.
- the fact that the momentum with which a charged particle approaches a junction is unchanged (as a vector) as the charged particle leaves the junction.
- the fact that there is no accumulation of charges at a junction.
In a meter bridge the point D is a neutral point (Figure).
- The meter bridge can have no other neutral point for this set of resistances.
- When the jockey contacts a point on meter wire left of D, current flows to B from the wire.
- When the jockey contacts a point on the meter wire to the right of D, current flows from B to the wire through galvanometer.
- When R is increased, the neutral point shifts to left.
The circuit in figure shows two cells connected in opposition to each other. Cell E1 is of emf 6V and internal resistance 2Ω; the cell E2 is of emf 4V and internal resistance 8Ω. Find the potential difference between the points A and B.
