In the given circuit, the current flowing through the resistance $20\ Omega$ $20\ \Omega$ is $0.3text A$ $0.3\text{ A}$, while the ammeter reads $0.9text A$ $0.9\text{ A}$. The value of $R_1$ $R_1$ is ________ $Omega$ $\Omega$.
The diagram displays a circuit consisting of three parallel branches containing R1, a 20 Ohm resistor, and a 15 Ohm resistor, with an ammeter in series.

Numerical Answer Type:

Enter a numerical value Answer: 30 to 30 +4 marks

Solution & Explanation

### Related Formula For parallel branches, the potential difference

V

across each branch is identical:

V = I_i R_i

According to Kirchhoff's Current Law, the total current

I_texttotal

is the sum of currents in all parallel branches:

I_texttotal = i_1 + i_2 + i_3

### Core Logic Analyzing the circuit diagram: * Branch 1: Current

i_1

through

20\ Omega

0.3text A

. * Branch 2: Contains

15\ Omega

resistor with current

i_2

. * Branch 3: Contains resistor

R_1

with current

i_3

. Since the branches are in parallel, they have the same potential difference

V_AB

V_AB = i_1 times 20\ Omega = 0.3text A times 20\ Omega = 6text V

### Step 1: Calculate Currents Current through the second branch (

15\ Omega

resistor) is:

i_2 = fracV_AB15\ Omega = frac6text V15\ Omega = 0.4text A

Total current read by the ammeter is

0.9text A

. Thus:

i_1 + i_2 + i_3 = 0.9text A

0.3text A + 0.4text A + i_3 = 0.9text A

0.7text A + i_3 = 0.9text A implies i_3 = 0.2text A

### Step 2: Calculate R1 Now use the voltage relation for the branch containing

R_1

i_3 times R_1 = V_AB

(0.2text A) times R_1 = 6text V

R_1 = frac60.2 = 30\ Omega

Current directions and node equations in parallel circuit for Q58 — The diagram displays a circuit consisting of three parallel branches containing R1, a 20 Ohm resistor, and a 15 Ohm resistor, with an ammeter in series.

### Pattern Recognition In parallel networks, finding the branch voltage is always the primary step. Once

V = 6text V

is established, the remaining branch currents are easily found using Ohm's Law and current conservation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Current Electricity

Reference Study Guides

Current Electricity

More Current Electricity Previous-Year Questions — Page 6

Q43 jee_main_2024_30_jan_morning Temperature Dependence of Resistivity

An electric toaster has resistance of

60Omega

at room temperature

(27^circmathrmC)

. The toaster is connected to a

220mathrmV

supply. If the current flowing through it reaches

2.75mathrmA

, the temperature attained by toaster is around: (if

alpha = 2times 10^-4 / ^circmathrmC

)

A. $694^circ mathrmC$ $694^{\circ} \mathrm{C}$
B. $1235^circmathrmC$ $1235^{\circ}\mathrm{C}$
C. $1694^circmathrmC$ $1694^{\circ}\mathrm{C}$
D. $1667^circmathrmC$ $1667^{\circ}\mathrm{C}$

Solution

### Related Formula

R = fracVI

R = R_0 (1 + alpha Delta T)

### Core Logic First, evaluate the final resistance

R_T

at the operating condition using Ohm's law. Second, plug the final resistance into the linear temperature dependence equation for resistance to solve for final temperature

T

. ### Step 1: Calculate Final Resistance Given

V = 220 mathrm~V

and

I = 2.75 mathrm~A

R_T = frac2202.75 = 80 \,Omega

### Step 2: Apply Temperature Equation We know

R_27 = 60 \,Omega

R_T = R_27 [1 + alpha (T - 27)]

80 = 60 [1 + 2 times 10^-4 (T - 27)]

frac8060 = 1 + 2 times 10^-4 (T - 27)

frac43 - 1 = 2 times 10^-4 (T - 27)

frac13 = 2 times 10^-4 (T - 27)

### Step 3: Solve for T

T - 27 = frac16 times 10^-4

T - 27 = frac100006 = 1666.67

T = 1666.67 + 27 approx 1693.67 Rightarrow 1694^circ mathrmC

### Pattern Recognition Always separate the final temperature

T

from

Delta T

. The most common error is forgetting to add back the initial reference temperature (

27^circmathrmC

) at the end. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Current Electricity

Q52 jee_main_2024_30_jan_morning Cells in Opposition and Terminal Voltage

Two cells are connected in opposition as shown. Cell

E_1

is of

8mathrmV

emf and

2Omega

internal resistance; the cell

E_2

is of

2mathrmV

emf and

4Omega

internal resistance. The terminal potential difference of cell

E_2

is:

Cells in Opposition and Terminal Voltage diagram for Q52 - JEE Main 2024 Morning — Two batteries pushing current against each other.

Numerical Answer. Answer: 6 to 6

Solution

### Related Formula

I = fracE_textnetR_texteq

V = E - Ir quad (textDischarging)

V = E + Ir quad (textCharging)

### Core Logic

Circuit with marked nodes for Kirchhoff analysis. — Two batteries pushing current against each other.

Because the cells are in opposition, the net EMF drives current from the higher potential cell (

8mathrmV

) to the lower potential cell (

2mathrmV

). Thus, the

2mathrmV

cell acts as a load and undergoes charging. ### Step 1: Calculate Total Current

I = frac8 - 22 + 4 = frac66 = 1 mathrm~A

### Step 2: Terminal Potential of Cell 2 Since cell

E_2

is being charged, its terminal potential difference is:

V_2 = E_2 + I r_2

Applying Kirchhoff's rule across cell

E_2

(from node C to B):

V_C - V_B = E_2 + I r_2 = 2 + (1)(4) = 6 mathrm~V

### Pattern Recognition When a smaller battery is forced backwards by a larger battery, it gets "charged". Consequently, its terminal voltage increases above its nominal EMF:

V = E + Ir

. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Current Electricity

Q39 jee_main_2024_31_jan_evening Meter Bridge

The resistance per centimeter of a meter bridge wire is

r

, with

X \, Omega

resistance in left gap. Balancing length from left end is at

40 text cm

with

25 \, Omega

resistance in right gap. Now the wire is replaced by another wire of

2r

resistance per centimeter. The new balancing length for same settings will be at

A. $20 text cm$ $20 \text{ cm}$
B. $10 text cm$ $10 \text{ cm}$
C. $80 text cm$ $80 \text{ cm}$
D. $40 text cm$ $40 \text{ cm}$

Solution

### Related Formula

fracR_textleftR_textwire-left = fracR_textrightR_textwire-right

### Core Logic For a meter bridge, the balancing condition is independent of the absolute resistance of the bridge wire as long as it is uniform. The ratio of the resistances in the gaps balances with the ratio of lengths.

Meter Bridge diagram for Q39 - JEE Main 2024 Evening

### Step 1: First Condition

fracXr ell_1 = frac25r (100 - ell_1)

Given

ell_1 = 40 text cm

fracXr times 40 = frac25r times 60 implies fracX40 = frac2560

### Step 2: Second Condition When replaced by a wire of

2r

per cm, the new lengths

ell_2

will satisfy:

fracX2r ell_2 = frac252r (100 - ell_2)

Notice that the

2r

terms cancel out entirely from both sides, leaving:

fracXell_2 = frac25100 - ell_2

### Step 3: Conclusion Since the ratio

X/25

remains identical, the balancing length ratio

ell / (100-ell)

also remains identical. Therefore,

ell_2 = ell_1 = 40 text cm

. ### Pattern Recognition Meter bridge balance point strictly depends on length ratio, NOT the specific resistivity or thickness of the wire (provided it is uniform). If external resistors don't change, the balance point never changes. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Current Electricity

Q43 jee_main_2024_31_jan_evening Power Dissipation

By what percentage will the illumination of the lamp decrease if the current drops by

20\%

A. $46\%$ $46\%$
B. $26\%$ $26\%$
C. $36\%$ $36\%$
D. $56\%$ $56\%$

Solution

### Related Formula Power (Illumination) is proportional to the square of the current for a constant resistance:

P = I^2 R

### Core Logic Initial power:

P_1 = I_1^2 R

If current drops by 20%, the new current is:

I_2 = I_1 - 0.2 I_1 = 0.8 I_1

New power:

P_2 = (0.8 I_1)^2 R = 0.64 I_1^2 R = 0.64 P_1

### Step 1: Calculate Percentage Change

Delta P \% = fracP_2 - P_1P_1 times 100\%

Delta P \% = frac0.64 P_1 - P_1P_1 times 100\%

Delta P \% = (0.64 - 1) times 100\% = -36\%

### Step 2: Final Statement The negative sign indicates a decrease. The illumination drops by

36\%

. ### Pattern Recognition For squared relations

y = x^2

, if

x

changes by a factor

k

(

0.8

y

changes by a factor

k^2

(

0.64

1 - 0.64 = 36\%

. This bypasses algebraic limits usually done for small changes (like

2Delta x / x

) since 20% is too large for approximation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Current Electricity

Q51 jee_main_2024_31_jan_evening Power in DC Circuits

In the following circuit, the battery has an emf of

2 text V

and an internal resistance of

frac23 \, Omega

. The power consumption in the entire circuit is ______ W.

Power in DC Circuits diagram for Q51 - JEE Main 2024 Evening — The image shows a DC circuit with multiple resistors in parallel/series connected to a 2V battery with 2/3 ohm internal resistance.

Numerical Answer. Answer: 3 to 3

Solution

### Related Formula

P = fracV^2R_eq

### Core Logic To find total power, collapse the entire external circuit and battery internal resistance into a single equivalent resistance

R_eq

across the battery's ideal terminals. ### Step 1: Equivalent Resistance Calculation Analyzing the diagram: The circuit simplifies to an equivalent resistance

R_eq

combining the parallel/series elements along with the internal resistance

r = 2/3 \, Omega

. The final simplified equivalent resistance of the entire system calculates to:

R_eq = frac43 \, Omega

### Step 2: Calculate Power

P = fracV^2R_eq

P = frac2^24/3

P = frac44/3 = 3 text W

### Pattern Recognition Whenever "entire circuit" power is asked, include the battery's internal resistance inside

R_eq

so you can use

P = E^2 / R_total

directly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Current Electricity

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