Types of Resistors: Important Questions

Resistors are available in different sizes, materials, and shapes. There are two basic types of resistors, namely, linear and nonlinear resistors. Linear resistors are those resistors, which values change with the applied voltage and temperature. In other words, in the linear resistors, the current value is directly proportional to the applied voltage. Nonlinear resistors are those resistors, where the current flowing through it does not change based on Ohm’s Law but, changes with a change in temperature or applied voltage. Moreover, if the change in body temperature influences the flowing current through a resistor, these resistors are called Thermistors. If the flowing current through a resistor changes with the applied voltages, it is called a Varistors or VDR (Voltage Dependent Resistors).

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Very Short Answer Question [1 mark Questions]

Ques: What do you mean by electric current? (1/5, Board Term 1,2017)

Ans: Electric current is defined as the amount of charge flowing through a particular area in unit time.

Ques: Define one ampere. (1/5, Board Term 1,2015)

Ans: One ampere can be constituted by the flow of one coulomb of charge per second. 1 A = 1 C s^-1

Ques: If a person has five resistors where each of value 1/5 Ω, then the maximum resistance that he can obtain by connecting them is
(a) 1 Ω
(b) 5 Ω
(c) 10 Ω
(d) 25 Ω (2020)

Ans: (a) The maximum resistance can be obtained by connecting them in series from a group of resistors. Thus,
R_s = 1/5+1/5+1/5+1/5+1/5 1 Ω

Ques: The maximum resistance which can be made using four resistors each of 2 Ω is
(a) 2 Ω
(b) 4 Ω
(c) 8 Ω
(d) 16 Ω (2020)

Ans: (c) A group of resistors can produce maximum resistance when they all are connected in series.
∴ R_s = 2 Ω + 2 Ω + 2 Ω + 2 Ω = 8 Ω

Ques: The maximum resistance which can be made using four resistors each of resistance 12 Ω is
(a) 2 Ω
(b) 1 Ω
(c) 2.5 Ω
(d) 8 Ω (2020)

Ans: (a) The maximum resistance can be produced from a group of resistors by connecting them in series.
Thus, R_s = 12 Ω + H 12 Ω + 12 Ω + 12 Ω = 2 Ω

Ques: Three resistors of 10 Ω, 15 Ω, and 5 Ω are connected in parallel. Find their equivalent resistance. (Board Term I, 2014)

Ans: Here, R₁ = 10 Ω, R₂ =15 Ω, R₃ = 5 Ω.
In parallel combination, equivalent resistance, (R_eq) is given by,

1/R_eq = 1/R₁ + 1/R₂ + 1/R₃

Hence, 1/R_eq = 1/10 + 1/15 + 1/5

1/R_eq = 3+2+6/ 30 = 11/30

R_eq = 30/11Ω = 2.73 Ω.

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Short Answer Question [2 marks Questions]

Ques: State Ohms law. (AI 2019)

Ans: It states that the potential difference V, across the ends of a given metallic wire in an electric circuit is directly proportional to the current flowing through it, provided its temperature remains the same. Mathematically,
V ∝ I
V = RI
where R is the resistance of the conductor.

Ques: A V-I graph for a nichrome wire is given below. What do you infer from this graph? Draw a labeled circuit diagram to obtain such a graph. (2020)

Ans: As the graph is a straight line, so it is clear from the graph that V ∝ I.

The shape of the graph obtained by plotting the potential difference applied across the conductor against the current flowing v. llmuigh il will be a straight line.
According to ohms law,

V = IR or R = VI
So, the slope of V’-/ graph at any point represents the resistance of the given conductor.

Ques: Show how can be the three resistors joined, each of resistance 9 Ω so that the equivalent resistance of the combination is (i) 13.5 Ω, (ii) 6 Ω. (2018)

Ans: (i) The resistance of the series combination is much higher than each of the resistances. A parallel combination of two 9 Ω resistors is equal to 4.5 Ω. We can get 13.5 Ω by combining 4.5 Ω and 9 Ω in series. So, to obtain 13.5 Ω, the combination is as shown in the figure below:

(ii) In order to get an equivalent resistance of 6 Ω, first we have to connect two 9 Ω resistors in series and then connect the third 9 Ω resistor in parallel to the series combination as shown in the below figure:

Ques: Calculate the ratio of equivalent resistance in two cases where the two identical resistors are first connected in series and then in parallel. (Board Term I, 2013)

Ans: Let the resistance of each resistor be R.
For series combination,
R_s = R₁ + R₂
So, R_s = R + R = 2R
For parallel combination,

1/R_p = 1/R₁ + 1/R₂ 

or,  = R_p = R₁R₂/(R₁ + R₂)

The required ratio will be = R_s/R_p = 2R/R/2 = 4:1

Ques: What are the fixed resistor types? 

Ans: Fixed resistors can be classified into three types. The first is a carbon composition resistor, which is the most common and has a resistive substance made of carbon clay and tin copper leads. The second type is metalized resistor, which, like constantan, is composed of 60% copper and 40% nickel, and magnesium material has a high level of resistivity. The third type of resistor is the wired wound type, which is created using a film deposition technique that incorporates a thick film of resistive material into an insulating substance.

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Long Answer Questions [3 marks Questions]

Ques: Three resistors of 3 Ω each are connected to a battery of 3 V as shown. Calculate the current drawn from the battery. (Board Term I, 2017)

Ans: As in the circuit diagram, two 3 Ω resistors are connected in series to form R₁; so R₁ = 3 Ω + 3 Ω = 6 Ω
And, R₁ and R₂ are in parallel combination, Hence, the equivalent resistance of the circuit (R_eq) is given by:

R_eq = 2 Ω
By applying Ohm’s law, V = IR
We get,
3 V = I × 2 Ω
or I = 3/2 A = 1.5 A
The current drawn from the battery is 1.5 A.

Ques: (a) If a 6 Ω resistance wire is doubled on itself, then find out the new resistance of the wire.
(b) Three 2 Ω resistors A, B, and C are connected in such a way that the total resistance of the combination is 3 Ω. Justify your answer by showing the arrangement of the three resistors. (2020)

Ans: (a) Given the resistance of the wire, R = 6 Ω
Let l be the length of the wire and A be its area of cross-section. Then,
R = ρl/A = 6 Ω
Now when the length is doubled, l’ = 2l and A’ = A2
∴ R’ = ρ(2l)/A/2=4ρl/A = 4 × 6 Ω = 24 Ω

(b)  As given the total resistance of the combination = 3 Ω
To get a total resistance of 3 Ω, the three resistors has to be connected as shown below:

Such that, 1/R_P=1/2+1/2 = 1
⇒ R_p = 1 Ω
and R_s = 2 Ω + 1 Ω = 3 Ω

Ques: Draw a schematic diagram of a circuit that comprises a battery of 3 cells of 2 V each, a combination of three resistors of 10 Ω, 20 Ω, and 30 Ω, that are connected in parallel, a plug key, and an ammeter, all connected in series. To find the value of the following use this circuit:
(a) Current through each resistor
(b) Total current in the circuit
(c) Total effective resistance of the circuit. (2020)

Ans: The circuit diagram is shown below:

(a) Given, voltage of the battery = 2V + 2V + 2V = 6 V

Current through 10 Ω resistance,
I₁₀ = V/R=6/10 = 0.6 A

Current through 20 Ω resistance,
I₂₀ = V/R=6/20 = 0.3 A

Current through 30 Ω resistance,
I₃₀ = V/R=6/30 = 0.2 A

(b) Total current in the circuit, 1= I₁₀ + I₂₀ + I₃₀
= 0.6 + 0.3 + 0.2 = 1.1 A

(c) Total resistance of the circuit,
1/R_P=1/10+1/20+1/30=11/60

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Very Long Answer Questions [5 marks Questions]

Ques: Current in each resistor, for the series combination of three resistors, establish the relation R = R₁ + R₂ + R₃ where the symbols have their usual meanings. Find out the equivalent resistance of the combination of three resistors of 6 Ω, 9 Ω and 18 Ω joined in parallel. (Board Term I, 2016)

Ans: 

The above figure shows the series combination of three resistors R₁, R₂ and R₃ connected across a voltage source of potential difference V.

Let current I is flowing through the circuit. Then,
V₁, V₂ and V₃ are the potential differences across resistors R₁, R₂ and R₃ respectively.

Since the total potential difference across a combination of resistors in a series is equivalent to the sum of potential differences across the individual resistors.
Therefore, v = v₁ + v₂ + v₃ …(i)

In series current through each resistor is the same.

Applying the Ohms law,
V₁ = IR₁, V₂ = IR₂ and V₃ = IR₁ ……..(ii)

If R_s is the equivalent resistance of the circuit, then
V = IR_s …(iii)
From the equations: (i), (ii) and (iii),
IR_s = IR₁ + IR₂ + IR₃

or, R_s = R₁ + R₂ + R₃

From here, we can conclude that when several resistors are joined in series, the resistance of the combination R_s equals the sum of their individual resistances,
R₁, R_2, and R₃

As given: R₁ = 6 Ω, R₂ = 9 Ω,
R₃ = 18 Ω are connected in parallel.
Equivalent resistance, R_eq, is given by:

Ques: Draw a labeled circuit diagram that shows three resistors R₁, R_2, and R₃ connected in series with a battery (E), a rheostat (Rh), a plug key (K), and an ammeter (A) by using standard circuit symbols. Also, use this circuit to show that the same current flows through every part of the circuit. Name two precautions you would observe while performing the experiment. (Board Term I, 2014)

Ans: 

Change the positions of the ammeter and note the reading of the ammeter each time. All the readings obtained are the same. So, the value of the current in the ammeter can be considered the same, independent of its position in the electric circuit. This means that in this circuit, which is a series combination, the current is the same in every part of the circuit.

The precautions one can observe while performing the experiment are:
(i) All the connections can be found neat and tight.
(ii) Ammeter is connected with the proper polarity, that is, the positive terminal of the ammeter should go to the positive terminal and the negative terminal of the ammeter to the negative terminal of the battery or cell used.

Ques: The network PQRS, shown in the circuit diagram, has batteries of 4 V and 5 V and negligible internal resistance. A milliammeter of 20 Ω resistance is connected between P and R. Calculate the reading in the milliammeter. (Comptt. All India 2012)

Ans: Applying loop rule to loop PQRP
-4 = 60(I – I₁) – 20 I₁ = 0
or – 4 = 60I – 60I₁ – 20I₁
or 20I₁ -15 I = 1 …[+ by 4 …(i)]

Applying loop Yule to loop PRSP, we get
-5 + 200 I + 20 I₁ = 0
4I₁ + 40 I = 1 …[+ by 5 …(ii)]

Therefore, the Reading of the milliammeter = 0.064 A.

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