Analytical Questions

 Critical Thinking based  Problem-Solving

Analytical questions typically involve critical thinking and problem-solving.

Circuit Analysis: A circuit with multiple resistors in series and parallel is given, one needs to determine the total resistance and current flowing through the circuit. Use knowledge of Ohm's law and the rules for combining resistors in series and parallel to find the solution.

Voltage Distribution: In a circuit with bulbs connected in parallel, calculate the current across each bulb if the total voltage supplied is known. Consider how the parallel connection affects voltage distribution. Also one needs to identify and explain what happens to the overall circuit when one component fails or is removed.

Power Dissipation: Given the resistances of different components in a circuit, calculate the power dissipated by each component. Consider all combination of the  the formulas for power (P = VI or P = I² R 0r P = V²/ R) and how the resistances affect power dissipation.

These types of questions require applying theoretical concepts, using formulas, and applying problem-solving skills to analyze and deduce outcomes or behaviors within a circuit or physics context.

1. 

A1, A2, A3 and A are ammeters connected in the circuit

B1, B2 and B3 are three identical bulbs
They all are connected to Voltage source as shown in Figure
When the three bulb are working good and glowing, the current recorded in Ammeter A is 6 A

Answer Following questions

a. Same amount of current will go through each Bulb.  And the value is 2 A .True or False

b. If the Bulb B3 is blown away, the bulb B1 and B2 will start glowing more. True or False

c. What will happen to the entire ammeter reading if Bulb B1 is blown away?

d. The current shown in Ammeter A remains even any bulb goes down. True or False

A1, A2, A3 and A are ammeters connected in the circuit
B1, B2 and B3 are three identical bulbs
They all are connected to Voltage source as shown in Figure
When the three bulb are working good and glowing, the current recorded in Ammeter A is 6 A

Answer Following questions

a. Same amount of current will go through each Bulb.  And the value is 2 A .True or False

b. If the Bulb B3 is blown away, the bulb B1 and B2 will start glowing more. True or False

c. What will happen to the entire ammeter reading if Bulb B1 is blown away?

d. The current shown in Ammeter A remains even any bulb goes down. True or False

Solution:

a. True. Since Bulb are identical and connected in parallel with Voltage. Same current will flow through each bulb. Since the total current is 6 A. Individual current will be 2 A

b. False. If the Bulb B3 is blown away, the potential difference across other bulb still remains same, So same current will flow and they will glow as it is. No change.

c. when Bulb B1 goes down, the current in that part become zero. So reading of Ammeter A1 becomes zero.

Reading of Ammeter A2 will remain same i.e. 2 A

Reading of Ammeter A3 will remain same i.e. 2 A

Reading of Ammeter A will be = 2 + 2= 4 A

d. As shown above, the reading of Ammeter A will change.

2. 

Solution:

Given:
Reading of ammeter A1​ is 1 ampere.

Let R1​ be the equivalent resistance along V1​.

So, R1​= 3 x 3 ÷ (3 + 3) ​= 1.5 Ω

So, reading of V1 is given by

V₁​ = I x R1 = 1×1.5 

     = 1.5 Volt

Let R2 be the equivalent resistance along V2.
1/R2 ​​= 1/3 + 1/3 + 1/3

        ​= 3/3​
R2    ​= 1 Ω
So, reading of V2​   = R₂ x A₁ = 1 ×1 = 1 V

Now, Let R3  ​be the equivalent resistance along A2​
R3​ = 3/4​ Ω
Equivalent resistance of the circuit is given by
R  =  R1​  + R2 ​+ R3

    ​= 1.5 + 1 + 0.75

    = 3.25 Ω
Since R1, R2, and R3 are connected in series. So, the relation between A1​and A3 is given by
A1​= A3​= 1 ampere
Reading of A2​is given by
i2​ = 1/4

    = 0.25 ampere

3. An electric lamp of 100 Ω, a toaster of resistance 50 Ω, and a water filter of resistance 500 Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as must current as all three appliances and what is the current through it?

Solution:

The electric iron will have equivalent resistance to the same as 100 Ω, 50 Ω and 500 Ω resistors are in parallel.
Let it be R Ω.

1/R = 1/100 + 1/50 + 1/500

       = (5 + 10 + 1) / 500

       = 16/500

  R  = 31.25 Ω.

Now, to find current

From ohm's law,

I = V / R

  = 220/31.25

  = 7.04 A

4. In the following given circuits, the heat produced in the resistor or combination of resistors
    connected to a 12 V battery will be -

a. Same in all the cases                    b. Minimum in case (i)

c. Maximum in case (ii)                   d. Maximum in case (iii)

Solution:

Case (i):

I = ^V/_R 

  = ¹²/₂ 

  = 6 A

H = I² R t

    = 6 x 6 x 2 t

    = 72 t Joule

Case (ii):

R = 2 + 2

    = 4 Ω

I = ^V/_R 

  = ¹²/₄ 

   = 3 A

H = I² R t

    = 3 x 3 x 4 t

    = 36 t Joule

Case (iii):

R = ^{2 x 2}/ _{2 + 2} 

    = 1 Ω

I = ^V/_R

   = ¹²/₁ 

   = 12 A

H = I² R t

    = 12 x 12 x 1 t

    = 144 t Joule

∴ option d is the correct answer.

5. In the given circuit, A, B, C, and D   are four lamps connected with a battery of 60 V.

Analyze the circuit to answer the following questions-

i.  What kind of combination are the lamps arranged in (series or parallel)

ii.  Explain with reference to your above answer, what are the advantages of the combination of lamps.

iii. Explain with proper calculations which lamp glows the brightest?

iv. Find out the total resistance of the circuit.

Solution:

 i. The lamps are in parallel connection.

 ii. Advantages of parallel combination over series combination:

1. If one appliance stops working or goes out of order, then all other appliances keep on working.
2. All appliances can be operated at the same voltage as the electric supply.
3. Different appliances have different requirements of current. This cannot be satisfied in series as the current remains the same in series.
4. The total resistance in a parallel circuit is decreased.
5. All devices can be operated independently with separate switches.

iii. The lamp with the highest power will glow the brightest.

     We know, P = V x I

     PA = 60 x 3 = 180 Watts

     PB = 60 x 4 = 240 Watts

     PC = 60 x 5 = 300 Watts

     PD = 60 x 3 = 180 Watts

     ∴ Bulb C will glow the brightest.

iv. To find the total resistance are –

      We know, P = V x I or V² /R

 R = V²/P

RA = 60 x 60 / 180 = 20 Ω

RB = 60 x 60 / 240 = 15 Ω

RC = 60 x 60 / 300 = 12 Ω

RD = 60 x 60 /180 = 20 Ω

Since all are in parallel, therefore the equivalent resistance is –

1/R = 1/RA + 1/RB + 1/RC + 1/RD

       = 1/20 + 1/15 + 1/12 + 1/20

       = (6 + 8 + 10 + 6) / 120

R   = 120/30 Ω

      = 4 Ω

Or R = Voltage Applied / Total Current

         = 60/(3 + 4 + 5 +3)

         = 60/15

        = 4 Ω

6. The diagram below is a schematic diagram of a household circuit. The house shown in the above diagram has 5 usable spaces where electrical connections are made. For this house, the mains have a voltage of 220 V and the net current coming from the mains is 22A.

a. What is the mode of connection to all the spaces in the house from the mains?

b. Spaces 5 and 4 have the same resistance and spaces 3 and 2 have respective resistances of 20
   Ω and 30 Ω. Space 1 has a resistance double that of space 5. What is the net resistance for
    space 5?

c. What is the current in space 3?

d. What should be placed between the main connection and the rest of the house’s electrical
     appliances to save them from accidental high electric current?

Solution:

a. All spaces are connected in parallel. All Electrical and Electronic devices works on rated voltage and current. 

b. Let the resistance of spaces 4 and 5 be R Ω

Resistance of Space 1 = 2R Ω

Resistance of Space 2 = 30 Ω

Resistance of Space 3 = 20 Ω

Current = 22 A

V = 220 V

Total resistance R𝑒𝑞 = V/I

                                 = 220/22

                                 = 10 Ω

Now, from parallel combination,

1/R𝑒𝑞 = 1/2R  + 1/30 + 1/20 + 1/R  + 1/R

 1/10   = 1/2R + 2/R + (20 x 30)/(20 + 30)     

  ⇒ ⁵/_2R = ¹/₁₀ – ¹/₃₀ + ¹/₂₀ 

              = ¹/₆₀

⇒ R      = 60 x ⁵/₂ 

             = 150 Ω      

c. Current in Space 3 (I₃₎ = ^V/_R3 

                                         = ²²⁰/₂₀ 

                                         = 11 A

d. Fuse or the Circuit Breaker.

7. In the circuit given below if the current reading in the ammeter A is 2 A, what would be the value of
     R₁?

Total resistance of the circuit = R𝐞𝐪

                                                = V/I 

                                                = 30/2 

                                                = 15 Ω

5 Ω, 10 Ω, and R1 are in parallel.

∴ 1/R𝐩 = 1/5 + 1/10 + 1/R1

⇒ 1/R𝐩 = (2R1 + R1 +10) /10R1

⇒ R𝐩 = 10R1/(3R1 + 10)

Now, 6 Ω, R𝐩, and 6 Ω are in series.

∴ R𝐞𝐪 = 6 + 6 + 10R1/3R1 + 10 

            = 12 + 10R1/3R1 + 10

     15   = 12 + 10R1/3R1 + 10

     3 (3R1 + 10) = 10R1

     R1 = 30 Ω

8. Calculate the total resistance of the circuit and find the total current in the circuit.

 Solution:

 Equivalent resistance of R3 and R4 (in series) = R3 + R4 

                                                                           = 6 + 4 

                                                                           = 10 Ω  (Say R5)

Equivalent resistance of R2 and R5 (in parallel) = R2 x R5/R2 + R5 

                                                                             = 10 x 10 /10 + 10 

                                                                             = 100/20 

                                                                             = 5 Ω  (Say R6)

Equivalent resistance of R1 and R6 (in series) = R1 + R6 

                                                                           = 7 + 5 

                                                                           = 12 Ω

Now, by Ohm’s law,

V = IR

I = V/R 

   = 24/12 

   = 2 A

9.Vinita and Ahmed demonstrated a circuit that operates the two headlights and the two sidelights of a car, in their school exhibition. Based on their demonstrated circuit, answer the following questions.

(i) State what happens when switch A is
     connected to

     a) Position 2

     b) Position 3

(ii) Find the potential difference across each
      lamp when lit.

(iii) Calculate the current

        a. In each 12 Ω lamp when lit.

        b. In each 4 Ω lamp when lit.

(iv) Show, with calculations, which type of lamp, 4.0 Ω or 12 Ω, has the higher power.

Solution:

(i) a. Switch A when connected to position 2 then 12 Ω lamps (only) gets on.

     b. Switch A when connected to position 2 then 4 Ω lamps (only) gets on.

(ii) 12 V for both sets of lamps and all of them are in parallel. In parallel combination, the same
       voltage is applied across each device.

(iii) a. 12 Ω lamps are on or lit when the wire is connected to position 2.

       Voltage across both 12 Ω lamps = 12 V.

       V = I R (Ohm’s law)

    ⇒ I = V/R

    ⇒ I = 12/12 

          = 1 A.

   b. 4 Ω lamps are on when the wire is connected to position 3.

       Voltage across both 4 Ω lamps = 12 V.

       V = I R (Ohm’s law)

    ⇒ I = V/R

                ⇒ I = 12/4 

                       = 3A

iv). 

      P  = V x I   or  V²/R

      All lamps are in parallel and hence the same V for all lamps.

      For 4 Ω lamps → P = ^12×12/₄ 

                                      = 36 W

     For 12 Ω lamps → P = ^12×12/₁₂

                                       = 12 W

     Hence 4 Ω lamps will have higher power.

10. A household uses the following electrical appliances:

                i. Refrigerator of rating 400 W for ten hours each day.

                ii. Two electric fans of rating 80 W each for twelve hours each day.

                iii. Six electric tubes of rating 18 W each for 6 hours each day.

           Predict the amount of electricity bill of the household for the month of June if the cost is Rs.
           3.00/unit.

Solution:

      June month is of 30 days.

We know, E = P x t

Electricity consumption in one day = 400 x 10 + 2 x 80 x 12 + 6 x 18 x 6

                                                         = 4000 + 1920 + 648

                                                         = 6568 Watt hour

                                                         = 6568/1000 kWh

                                                         = 6.568 kWh

                                                         = 6.568 units.

Electricity consumption in one month = 6.568 x 30

                                                              = 197.04 units

∴ Total cost = Rs. 197.04 x 3

                    = Rs. 591.12