A power supply of electromotive force (e.m.f) 50V and negligible internal resistance connected in series with resistors of resistance 100\Omega and 5\Omega as shownA voltmeter measures the potential difference (p.d) across the 5\Omega resistor and an ammeter measures the current in the circuit.What are suitable ranges for the ammeter and for the voltmeter?

It is given that E= 50\ V R_eq = R_! +R_2 \implies R_eq = (100+5) \Omega = 105 \Omega So, Current I = \dfrac{50}{105}A \implies I = 0.48A P.d. across 5\ \Omega =I \times 5\Omega \implies V = 0.48 \times 5 \implies V = 2.4 V Since the current and voltage value fall in the range of ammeter and voltmeter given in option D.So, option D is correct.

In the circuit shown, the current in the 1\ \Omega resistor is:

The correct answer is option (C). Hint: Divide the circuit into two loops and apply Kirchhoff’s voltage loop formula. Step 1: Division of the circuit. The circuit can be divided into two loops as shown below: Step 2: Apply KVL in loop 1 . 9−2i−1\left(i−i_1\right)-3i=0 \Rightarrow 6i−i_1=9 …(i) Step 3: Apply KVL in loop 2 . 6−3i_1+1(i-i_1)=0 \Rightarrow -i+4i_1=6 …(ii) Step 4: Solve equations (i) and (ii) We multiply equation (i) by 4 4(6i-i_1)=9\times4 \Rightarrow24i-4i_1=36 …(iii) Adding equations (ii) and (iii), we get, 23i=42 \Rightarrow i=1.826A Putting the values of i in equation (i), we get, 6\times1.826-i_1=9 \Rightarrow i_1=1.956A Current through 1\Omega resistor is i-i_1=-0.13A (Minus sign indicates that the current flows from Q to P ) Hence, the current through 1\Omega resistor is 0.13A from Q to P . The correct answer is option (C).

Current Electricity - Problem solving tips

Q: A wire of uniform thickness with a resistance of 27\, \Omega is cut into three equal pieces and they are joined in parallel. The equivalent resistance of the parallel combination will be :

In this case, a wire of uniform thickness with a resistance of 27 is cut into three equal pieces and they are joined in parallel. As we know that the resistance is directly proportional to the length of the wire, R\propto L , so when the wire is cut into 3, the resistance of the cut wire is reduced to one third of the total resistance. That is, the resistance of each cut wire becomes 27/3= 9 ohms.Now, when these 3 wires are connected in a parallel connection, the total resistance is given as \frac { 1 }{ 9 } +\frac { 1 }{ 9 } +\frac { 1 }{ 9 } =\frac { 1 }{ 0.333 } =\quad 3\quad ohms .Hence, the total resistance is now 3 ohms only.

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Current Electricity

- Want to score better marks in exams? Have a look at some problem solving tips.

Tip 1:
Voltage division in a series circuit:

First of all, find the equivalent (total) resistance of the circuit by simply adding up all the series resistors.
Now the current drawn can be obtained by using the net resistance and the supply voltage.
Now as the current remains constant for all the resistors, the voltage across any resistance can be obtained using Ohm's law.

Considering a shortcut, the voltage across any particular resistance can be obtained by using the formula:

V_{R} = V_{s} \frac{R}{R _{t o t a l}}

Where

V_{R}

is the voltage drop across particular resistance,

R

is the resistance, and

V_{s}

is the voltage of the source.

A power supply of electromotive force (e.m.f)

50 V

and negligible internal resistance connected in series with resistors of resistance

100 Ω

and

5 Ω

as shown

A voltmeter measures the potential difference (p.d) across the

5 Ω

resistor and an ammeter measures the current in the circuit.

What are suitable ranges for the ammeter and for the voltmeter?

ammeter range

/ A : 0 - 0.1

voltmeter range

/ V : 0 - 1

ammeter range

/ A : 0 - 0.1

voltmeter range

/ V : 0 - 3

ammeter range

/ A : 0 - 1.0

voltmeter range

/ V : 0 - 1

ammeter range

/ A : 0 - 1.0

voltmeter range

/ V : 0 - 3

View solution

Tip 2:
Solving numerical based on Kirchhoff's Voltage Law:

For all independent loops, consider a loop current in the loop in a clockwise (or counter-clockwise) direction. These loop currents will be the unknown variables that we need to determine.
Apply KVL around each of the loops in the same clockwise direction to obtain the respective loop equations. While calculating the voltage drop across each resistor shared by two loops, both loop currents (in opposite positions) should be considered.
Solve all the equations obtained for the loops and determine the unknown currents.

In the circuit shown, the current in the

1 Ω

resistor is:

1.3 A

, from

P

Q

0 A

0.13 A

, from

Q

P

0.13 A

, from

P

Q

View solution

Tip 3:
Solve the parallel combination of equal resistances.

In the parallel combination of equal resistances, the net resistance is obtained as:

\frac{1}{R _{e q}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \frac{1}{R _{3}} + . . .

When

R_{1}, R_{2}, R_{3}, . .

are same, then the equation gets reduced as follow:

For

n

equal resistors in parallel, equivalent resistance is:

R_{e q} = \frac{R}{n}

A wire of uniform thickness with a resistance of

27 Ω

is cut into three equal pieces and they are joined in parallel. The equivalent resistance of the parallel combination will be :

3 Ω

9 Ω

27 Ω

49 Ω

View solution