Current Division MCQ Quiz - Objective Question with Answer for Current Division - Download Free PDF

Concept:

Current division rule:

When 'n' resistors are connected in parallel across a current source, then-current is divided as follows

\({I_{R1}} = I\left[ {\frac{{{\frac1{R_1} }}}{{\frac1{R_1} + \frac1{R_2}+ ~......~+\frac1{R_n}}}} \right]\)

When two resistors are connected in parallel, then-current is divided as follows

By using current division rule,

The current flows through the resistor R1 is \({I_{R1}} = I\left( {\frac{{{R_2}}}{{{R_1} + {R_2}}}} \right)\)

The current flows through the resistor R2 is \({I_{R2}} = I\left( {\frac{{{R_1}}}{{{R_1} + {R_2}}}} \right)\)

Calculation:

\(i_m = 5 \times \frac{10}{40+10} = 1~A\)

Since i_m is in the opposite direction it is taken in - 1 A.

Concept

By Ohm's Law:                                              

V = I × R

In parallel, the voltage remains constant.

Since the resistance is proportional to the length (R α l)

∴ \(I\space α\space {1\over R}\space α \space {1\over l}\)

where, I = Current

R = Resistance

l = Length

Calculation
 
The equivalent resistance for the circuit is:

\(R=R_1||R_2||R_3\)

\(R=({1×2\over 1+2})||4\)

\(R=({2\over 3})||4\)

\(R={{2\over 3}× 4\over {2\over 3}+4}=0.571\Omega \)

  
The voltage (V) is given by:

V = 10 × 0.571 = 5.71 volts

In parallel, the voltage remains the same.

\(I_1={5.71\over 1}=5.71 A\)

\(I_2={5.71\over 2}=2.85A\)

\(I_3={5.71\over 4}=1.42 A\)

The correct answer is option 1): 

Concept:

Current division rule

Current through resistor R₁,\(I_1=I×\frac{R_2}{R_1+R_2}\)

Current through resistor R_2,\(I_2=I×\frac{R_1}{R_1+R_2}\)

When two resistors connected in R1,  R2 series 

The total resistance is  R1 + R2

When two resistors connected in R1,  R2 Parallel

The total resistance is ( \(\frac{1}R_1 + \frac{1}{R_2}\))^-1

Calculation:

The given circuit is

Total resistance = 4 + ( \(\frac{1}{10} + \frac{1}{20}\))^-1

=  4+ 6.667

= 10.667

I = \(V\over R\)

= 1.87494 A

Current through resistor 20 ,\(I_2=I×\frac{R_1}{R_1+R_2}\)

= 1.87494 × \(10 \over 30\)

= 0.625 A

The correct answer is option 3):(32 mA)

Concept:

When two resistors R₁ and R₂ are connected in series the total resistance 

R_t = R1 +R2

When two resistors R1 and R2 are connected in parallel the total resistance will be

R_t = (\(1\over R_1\) + \(1\over R_2\))^-1

Current Division method

Current through R₁ \(I_1=I×\frac{R_2}{R_1+R_2}\)

Current through R₂ \(I_2=I×\frac{R_1}{R_1+R_2}\)

By  ohms law 

V = IR 

where I is the current R is the resistor

Calculation:

Given

The circuit  can be simplified as

R₁ = 500 Ω 

R₂ = 150 Ω + 600 Ω = 750 Ω 

Total resistance = ( \(1\over 500\) + \(1 \over 750\))^-1+ 200

= 500 Ω 

I = \(40 \over 500\)

= 0.08 A

\(I_2=I×\frac{R_1}{R_1+R_2}\) = i

= 0.08 × \(500 \over 1250\)

= 0.032 A

= 32mA

The correct answer is option 2):(-1 A)

Concept:

When two resistors R1 and R2 are connected in series the total resistance 

Rt = R1 +R2

When two resistors R1 and R2 are connected in parallel the total resistance will be

Rt = (\(1\over R_1\) + \(1\over R_2\))-1

Current Division method

Current through R1 \(I_1=I×\frac{R_2}{R_1+R_2}\)

Current through R2 \(I_2=I×\frac{R_1}{R_1+R_2}\)

By  ohms law 

V = IR 

where I is the current R is the resistor

Calculation:

The given circuit is 

Current through R2 \(I_2=I×\frac{R_1}{R_1+R_2}\)

= 5 × \(10\over 50\)

= 1 A

the direction of the current is reversed so I_m = -1 A

Given V = 12 V

Applying Ohms Law at 4 Ω resistor, we get:

12 = I × 4

I = 3 A

The circuit is redrawn as shown:

Using current division, we get:

\(I_1 =3\times \frac{R_1}{R_1+R_2}\)

\(I_1 =3\times \frac{8}{8+4}=2~A\)

So the ammeter will read 2 A.

Given V_s = 150 V, R₁ = 50 Ω, R₂ = 25 Ω

Since voltage parameter in parallel connected elements is same.

V_s = V₁ = V₂ = 150 V

∴ i₁ = V₁ / R₁ = 150 / 50 = 3 A

Since the direction of the arrow mentioned is in opposite direction to the actual current flow i₂ is considered as negative.

i₂ = -V₂ / R₂ = -150 / 25 = - 6 A

Given: 

The voltage drop across the 5 Ω resistors is 15 V

15 = 5I ⇒ I = 3 A

I will divide into I₁ and I₂

By current division 

\(I_1=\frac{5+5}{5+5+10}I=\frac{10}{20}× 3= 1.5\;A\)

\(I_2=\frac{10}{5+5+10}I=\frac{10}{20}× 3= 1.5\;A\)

Voltage across AC = V_AC = I₁ × 10 = 1.5 × 10 = 15 V

E = I(5 + 10) + VAC = 3 × 15 + 15 = 60 V

The EMF of the battery is 60 V.

Additional Information

Current division rule

By using current division rule,

The current flows through the resistor R­1 = I1 = \(\frac{{I\left( {{R_2}} \right)}}{{{R_1} + {R_2}}}\)

The current flows through the resistor R­2 = I2 = \(\frac{{I\left( {{R_1}} \right)}}{{{R_1} + {R_2}}}\)

Concept:

Current division rule

Consider the below circuit to understand the current division rule before solving the actual problem

By using the current division rule,

The current flows through the resistor R­1 = I1 = \(\frac{{I\left( {{R_2}} \right)}}{{{R_1} + {R_2}}}\)

The current flows through the resistor R­2 = I2 = \(\frac{{I\left( {{R_1}} \right)}}{{{R_1} + {R_2}}}\)

Calculation:

From concept

\(I_4 =\frac{{I\left( {{R_3}} \right)}}{{{R_3} + {R_4}}}\)

\(3 =\frac{{I\left( {{16}} \right)}}{{{16} + {8}}}\)

I = 4.5 A

Voltage across R₄ is V4

V₄ = I4 × R4 = 3 × 8 = 24 V

R₃ and R4 are in parallel so V₃ = V₄ = 24 V

V₁ = I × R₁ = 4.5 × 8 = 36 V

V2 = I × R2 = 4.5 × 6 = 27 V

From KVL,

E = V1 + V2 + V4 = 24 + 36 + 27

= 87 V

Concept: 

Current division rule:

When 'n' resistors are connected in parallel across a current source, then-current is divided as follows

\({I_{R1}} = I\left[ {\frac{{{\frac1{R_1} }}}{{\frac1{R_1} + \frac1{R_2}+ ~......~+\frac1{R_n}}}} \right]\)

When two resistors are connected in parallel, then-current is divided as follows

By using current division rule,

The current flows through the resistor R1 is \({I_{R1}} = I\left( {\frac{{{R_2}}}{{{R_1} + {R_2}}}} \right)\)

The current flows through the resistor R2 is \({I_{R2}} = I\left( {\frac{{{R_1}}}{{{R_1} + {R_2}}}} \right)\)

Application;

Let R₁ = 20 Ω, R₂ = 60 Ω, I = 40 A

\({I_{R2}} = I\left( {\frac{{{R_1}}}{{{R_1} + {R_2}}}} \right)\)

I_R2 = I_60Ω = 40 × (20 / 80) 

I_60Ω = 10 A

Concept:

Different connections of resistance.

1.Series Connection.

R_eq = R₁ + R₂ + R₃

If N resistance is connected in series then, 

R_eq = R₁ + R₂ + R₃

If N resistance is connected in series then, 

R­_eq = R₁ + R₂ + R₃ + ------ + R_N-1 + R_N = \( \mathop \sum \limits_{i = 1}^N {R_i}\)

2 Parallel Connection

\( \frac{1}{{{R_{eq}}}} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}}\)

• If N resistance are connected in parallel then,

 \( \frac{1}{{{R_{eq}}}} = \;\frac{1}{{{R_1}}} + \frac{1}{{{R_2}}} + \frac{1}{{{R_3}}} + - - - - + \frac{1}{{{R_{N - 1}}}} + \frac{1}{{{R_N}}} = \;\mathop \sum \limits_{i = 1}^N \frac{1}{{{R_i}}}\)

Current Division Rule:

The current in any of the parallel branches is equal to the ratio of opposite branch resistance to the total resistance, multiplied by the total current.

\( {I_1} = \;\frac{{{R_2}}}{{{R_1} + {R_2}}}I\;\;\;\& \;\;\;{I_2} = \frac{{{R_1}}}{{{R_1} + {R_2}}}I\)

Calculation:

Here, V = 10 V

R_a = 2 Ω

R_b = 8 Ω  

R_c = 5 Ω

R_x = R_a + R_b = 2 + 8 = 10 Ω

R_y = R_c + R = (5 + R) Ω 

R_eq = R_x || R_y =\( \frac{{{\rm{RxRy\;}}}}{{{\rm{Rx}} + {\rm{Ry}}}} = \;\frac{{10\left( {5 + R} \right)}}{{10 + 5 + R}} = \;\frac{{10\left( {5 + R} \right)}}{{15 + R}}\)

Since, V = IR_eq

_{\( 10 = 2.55 \times \frac{{10\left( {5 + R} \right)}}{{15 + R}}\)}

\( \;\;\frac{{15 + R}}{{5 + R}} = 2.55 \)

15 + R = 12.75 + 2.55R

1.55R = 2.25

R = 1.45 Ω

So R_y = 5 + R = 5 + 1.45 = 6.45 Ω

Now, through current division Rule 

\( \;\;{I_1} = \;\frac{{{R_y}}}{{{R_x} + {R_y}}} \times I\)

 =\( \frac{{6.45}}{{10 + 6.45\;}}\; \times 2.55\)

= \( \frac{{6.45 \times 2.55}}{{16.45}} = 1\;A\)

\( {I_2} = \;\frac{{{R_x}}}{{{R_x} + {R_y}}}I\)

= \( \frac{{10}}{{10 + 6.45}}2.55\)

= \( \frac{{10 \times 2.55}}{{16.45}} = 1.55\;A\)

Important Points

Voltage Division Rule:

Voltage across a resistor in a series circuit is equal to the value of that resistor times the total voltage across the series elements divided by the total resistance of the series elements.

\( {V_{R1}} = \;\frac{{{R_1}}}{{{R_1} + {R_2}}}{V_s}\;\;\& \;\;{V_{R2}} = \;\frac{{{R_2}}}{{{R_1} + {R_2}}}{V_s}\)

Shortcut Trick:

Resistance R_x = 10 Ω

The voltage across R_x is 10 V

So current \( {I_1} = \;\frac{V}{{{R_x}}} = \;\frac{{10}}{{10}} = 1\;A\)

From the 4 options given below, only option (2) contains I₁ = 1 Amp.

So option (2) is the correct option.

Concept:

Current division rule:

When 'n' resistors are connected in parallel across a current source, then-current is divided as follows

\({I_{R1}} = I\left[ {\frac{{{\frac1{R_1} }}}{{\frac1{R_1} + \frac1{R_2}+ ~......~+\frac1{R_n}}}} \right]\)

When two resistors are connected in parallel, then-current is divided as follows

By using current division rule,

The current flows through the resistor R1 is \({I_{R1}} = I\left( {\frac{{{R_2}}}{{{R_1} + {R_2}}}} \right)\)

The current flows through the resistor R2 is \({I_{R2}} = I\left( {\frac{{{R_1}}}{{{R_1} + {R_2}}}} \right)\)

Calculation:

\(i_m = 5 \times \frac{10}{40+10} = 1~A\)

Since i_m is in the opposite direction it is taken in - 1 A.

Concept

By Ohm's Law:                                              

V = I × R

In parallel, the voltage remains constant.

Since the resistance is proportional to the length (R α l)

∴ \(I\space α\space {1\over R}\space α \space {1\over l}\)

where, I = Current

R = Resistance

l = Length

Calculation
 
The equivalent resistance for the circuit is:

\(R=R_1||R_2||R_3\)

\(R=({1×2\over 1+2})||4\)

\(R=({2\over 3})||4\)

\(R={{2\over 3}× 4\over {2\over 3}+4}=0.571\Omega \)

  
The voltage (V) is given by:

V = 10 × 0.571 = 5.71 volts

In parallel, the voltage remains the same.

\(I_1={5.71\over 1}=5.71 A\)

\(I_2={5.71\over 2}=2.85A\)

\(I_3={5.71\over 4}=1.42 A\)

Concept:

By using the current division rule,

The current flows through the resistor R­1 = I1 = \(\frac{{I\left( {{R_2}} \right)}}{{{R_1} + {R_2}}}\)

The current flows through the resistor R­2 = I2 = \(\frac{{I\left( {{R_1}} \right)}}{{{R_1} + {R_2}}}\)

Calculation:

The given circuit can be reduced to,

Now, by using the current division rule, the current flowing through the 6 Ω resistor is

\(i = 12\left( {\frac{3}{{3 + 6}}} \right) = 4\;A\)

Concept:-

KVL: Kirchhoff’s Voltage Law (KVL) states that the sum of all the voltages around a loop is equal to zero.

KCL: Kirchhoff’s Current Law (KCL) states that the algebraic sum of all the current entering or exiting a node is always zero.

Current division: The distribution of total current in between the parallel branches of a divider circuit is known as current division.

Equation:

Current \({{\rm{I}}_1} = \frac{{{{\rm{R}}_2}}}{{{{\rm{R}}_1} + {{\rm{R}}_2}}}{{\rm{I}}_{\rm{T}}}\) ---(1)

Where I_T is the total current

I₁ and I₂ are the currents in branch 1 and branch 2 respectively.

Calculations:-

Given,

Voltage drop across 8 Ω resistor = 20 V

From ohm’s law equation

V = IR

∴\(\;{\rm{I}} = \frac{{\rm{V}}}{{\rm{R}}}\)

So, \({\rm{I}} = \frac{{20}}{8}{\rm{A}}\)  = 2.5 A    ----(2)

Using current division equation from  1 

\({{\rm{I}}_1} = \frac{{15 + 13}}{{15 + 13 + 11}}2.5{\rm{\;A}}\)

I₁ = 1.794 A

Now applying KVL in loop 1,

-E + 20 + 11 × I + 11 × I₁ = 0    

⇒ E = 20 + 11 × 2.5 + 11 × 1.794     (putting values of I and I₁)

⇒ E = 67.3 V

So the EMF of the battery shown is 67.3 Volts.