The heat generated in 1 minute between points A and B in the given circuit, when a battery of 9V with internal resistance of 1 Omega$1 \Omega$ is connected across these points is ____ J.
A bridge network with resistors 1 ohm, 2 ohms, 2 ohms, and 4 ohms forming the arms.
Numerical Answer Type:
Enter a numerical valueAnswer: 1080 to 1080+4 marks
Solution & Explanation
### Related Formula
R_texteq = fracR_1 R_2R_1 + R_2 text (parallel)$$R_{\text{eq}} = \frac{R_1 R_2}{R_1 + R_2} \text{ (parallel)}$$H = I^2 R t$H = I^2 R t$I = fracVR_texteq + r$$I = \frac{V}{R_{\text{eq}} + r}$$
### Core Logic
The circuit is a balanced Wheatstone bridge between A and B, because the ratio of adjacent arms is frac12 = frac24$\frac{1}{2} = \frac{2}{4}$. Thus, the middle 1\,Omega$1\,\Omega$ resistor (if there is one connecting the middle nodes) is ineffective.
A bridge network with resistors 1 ohm, 2 ohms, 2 ohms, and 4 ohms forming the arms.
The equivalent resistance across AB (R_AB$R_{AB}$) involves the upper branch (1+2 = 3\,Omega$1+2 = 3\,\Omega$) and the lower branch (2+4 = 6\,Omega$2+4 = 6\,\Omega$) in parallel.
A bridge network with resistors 1 ohm, 2 ohms, 2 ohms, and 4 ohms forming the arms.R_AB = frac3 times 63 + 6 = frac189 = 2\,Omega$$R_{AB} = \frac{3 \times 6}{3 + 6} = \frac{18}{9} = 2\,\Omega$$
### Step 1: Calculate Current and Heat
Total resistance in the circuit = R_AB + r_textinternal = 2 + 1 = 3\,Omega$R_{AB} + r_{\text{internal}} = 2 + 1 = 3\,\Omega$.
Total current from the battery i = frac9textV3\,Omega = 3textA$i = \frac{9\text{V}}{3\,\Omega} = 3\text{A}$.
The heat generated specifically between points A and B in 1 minute (60 seconds) is:
H_AB = i^2 R_AB t = (3)^2 times 2 times 60$$H_{AB} = i^2 R_{AB} t = (3)^2 \times 2 \times 60$$H_AB = 9 times 120 = 1080text J$$H_{AB} = 9 \times 120 = 1080\text{ J}$$
### Pattern Recognition
Check for balanced Wheatstone bridge first. Then use H = i^2 R t$H = i^2 R t$ strictly with the equivalent resistance of just the section AB$AB$ to find heat specifically generated across AB$AB$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
Keywords:#bridge circuit#resistors#heat generated
More Current Electricity Previous-Year Questions — Page 5
Q59jee_main_2024_27_jan_morningCapacitor in DC Circuit
The charge accumulated on the capacitor connected in the following circuit is ______ mutextC$\mu\text{C}$. (Given C = 150\ mutextF$C = 150\ \mu\text{F}$).
The layout shows an integrated network of resistors labeled R1 to R4 powered by a 10V DC source loop with a branch housing a 150 uF capacitor element bridging distinct structural potential nodes.
Numerical Answer.Answer: 400 to 400
Solution
### Related Formula
Q = C cdot Delta V_C$$Q = C \cdot \Delta V_C$$
### Core Logic
In steady state, no current flows through the capacitor branch. Analyze potential levels at node loops using standard Kirchhoff mesh values mapped across the core loop resistors:
V_A + frac103(1) - 6(1) = V_B$$V_A + \frac{10}{3}(1) - 6(1) = V_B$$
### Step 1: Compute node potential difference
V_A - V_B = 6 - frac103 = frac83text V$$V_A - V_B = 6 - \frac{10}{3} = \frac{8}{3}\text{ V}$$
### Step 2: Evaluate accumulated charge
Q = C(V_A - V_B) = 150\ mutextF times frac83text V = 50 times 8 = 400\ mutextC$$Q = C(V_A - V_B) = 150\ \mu\text{F} \times \frac{8}{3}\text{ V} = 50 \times 8 = 400\ \mu\text{C}$$
### Pattern Recognition
Steady state capacitor loops act as open electrical cuts, transforming active mesh equations into simple layout node checks.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
Q32jee_main_2024_29_jan_morningElectric Current and Charge
The electric current through a wire varies with time as I = I_0 + beta t$I = I_0 + \beta t$, where I_0 = 20 mathrm~A$I_0 = 20 \mathrm{~A}$ and beta = 3 mathrm~A / s$\beta = 3 \mathrm{~A / s}$. The amount of electric charge crossed through a section of the wire in 20 mathrm~s$20 \mathrm{~s}$ is:
A.80 mathrm~C$80 \mathrm{~C}$
B.1000 mathrm~C$1000 \mathrm{~C}$
C.800 mathrm~C$800 \mathrm{~C}$
D.1600 mathrm~C$1600 \mathrm{~C}$
Solution
### Related Formula
The relationship between current, charge, and time is given by:
I = fracdqdt implies q = int I \, dt$$I = \frac{dq}{dt} \implies q = \int I \, dt$$
### Core Logic
Given the current variation:
I = I_0 + beta t$$I = I_0 + \beta t$$
Substituting the given values, I_0 = 20 mathrm~A$I_0 = 20 \mathrm{~A}$ and beta = 3 mathrm~A/s$\beta = 3 \mathrm{~A/s}$:
I = 20 + 3t$I = 20 + 3t$
Thus:
fracdqdt = 20 + 3t$$\frac{dq}{dt} = 20 + 3t$$
### Step 1: Integrate to Find Charge
To find the total charge crossing a section of wire from t = 0$t = 0$ to t = 20 mathrm~s$t = 20 \mathrm{~s}$:
int_0^q dq = int_0^20 (20 + 3t) \, dt$$\int_{0}^{q} dq = \int_{0}^{20} (20 + 3t) \, dt$$q = left[ 20t + frac3t^22 right]_0^20$$q = \left[ 20t + \frac{3t^2}{2} \right]_{0}^{20}$$q = 20(20) + frac3(20)^22$$q = 20(20) + \frac{3(20)^2}{2}$$q = 400 + frac3(400)2 = 400 + 600 = 1000 mathrm~C$$q = 400 + \frac{3(400)}{2} = 400 + 600 = 1000 \mathrm{~C}$$
Therefore, the charge crossed is 1000 mathrm~C$1000 \mathrm{~C}$.
### Pattern Recognition
Whenever current is given as a function of time I(t)$I(t)$, the total charge is simply the area under the I-t$I-t$ curve, which mathematically corresponds to the definite integral int_t_1^t_2 I(t) \, dt$\int_{t_1}^{t_2} I(t) \, dt$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
Q57jee_main_2024_29_jan_morningKirchhoff's Laws and RC Circuits
A 16 \, Omega$16 \, \Omega$ wire is bend to form a square loop. A 9V battery with internal resistance 1 \, Omega$1 \, \Omega$ is connected across one of its sides. If a 4 \, mu mathrmF$4 \, \mu \mathrm{F}$ capacitor is connected across one of its diagonals, the energy stored by the capacitor will be fracx2 \, mu mathrmJ$\frac{x}{2} \, \mu \mathrm{J}$. where x = $x = $ _________.
Numerical Answer.Answer: 81 to 81
Solution
### Related Formula
Energy stored (U$U$) by a capacitor of capacitance C$C$ under steady state voltage V_C$V_C$:
U = frac12 C V_C^2$$U = \frac{1}{2} C V_C^2$$
### Core Logic
A wire of resistance 16 \, Omega$16 \, \Omega$ is bent into a square, so each of the 4 sides has a resistance of:
R_textside = frac164 = 4 \, Omega$$R_{\text{side}} = \frac{16}{4} = 4 \, \Omega$$
The battery is connected across one side. Let this side have resistance 4 \, Omega$4 \, \Omega$. The remaining three sides are connected in series, creating a parallel branch with a combined resistance of:
R_textseries = 4 + 4 + 4 = 12 \, Omega$$R_{\text{series}} = 4 + 4 + 4 = 12 \, \Omega$$Circuit network layout showing the square loop resistance divisions and diagonal capacitor connection for Q57
### Step 1: Calculate Equivalent Resistance and Circuit Current
The equivalent external resistance of the parallel loop branches is:
R_p = frac12 times 412 + 4 = frac4816 = 3 \, Omega$$R_p = \frac{12 \times 4}{12 + 4} = \frac{48}{16} = 3 \, \Omega$$
Including the battery's internal resistance (1 \, Omega$1 \, \Omega$), the total line current I$I$ leaving the battery is:
I = fracVR_p + r = frac93 + 1 = frac94 mathrm~A$$I = \frac{V}{R_p + r} = \frac{9}{3 + 1} = \frac{9}{4} \mathrm{~A}$$
### Step 2: Find Current through the Main Branches
Using the current divider rule, the current I_1$I_1$ flowing through the longer 12 \, Omega$12 \, \Omega$ branch is:
I_1 = I times frac412 + 4 = frac94 times frac416 = frac916 mathrm~A$$I_1 = I \times \frac{4}{12 + 4} = \frac{9}{4} \times \frac{4}{16} = \frac{9}{16} \mathrm{~A}$$
### Step 3: Find Potential Difference across the Diagonal
In a steady state, the capacitor acts as an open circuit. Let the diagonal link nodes be A$A$ and B$B$. The path contains two sides of the 12 \, Omega$12 \, \Omega$ branch (total resistance = 8 \, Omega$8 \, \Omega$):
V_C = V_A - V_B = I_1 times 8 = frac916 times 8 = frac92 mathrm~V$$V_C = V_A - V_B = I_1 \times 8 = \frac{9}{16} \times 8 = \frac{9}{2} \mathrm{~V}$$
### Step 4: Compute Stored Energy and Extract x
The stored energy U$U$ is:
U = frac12 times (4 \, mu mathrmF) times left(frac92right)^2 = frac12 times 4 times frac814 = frac812 \, mu mathrmJ$$U = \frac{1}{2} \times (4 \, \mu \mathrm{F}) \times \left(\frac{9}{2}\right)^2 = \frac{1}{2} \times 4 \times \frac{81}{4} = \frac{81}{2} \, \mu \mathrm{J}$$
Comparing this directly with the expression fracx2 \, mu mathrmJ$\frac{x}{2} \, \mu \mathrm{J}$:
x = 81$x = 81$
Therefore, the value of x$x$ is 81$81$.
### Pattern Recognition
Remember that capacitors act as standard open circuits when reaching steady-state DC conditions. Simply calculate the node potentials across the bridge connection points using standard current distribution laws first, then execute the energy equation.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
Class 12 Physics: Electrostatic Potential and Capacitance
Q48jee_main_2024_30_january_eveningPower Dissipation in Resistors
When a potential difference V$V$ is applied across a wire of resistance R$R$, it dissipates energy at a rate W$W$. If the wire is cut into two halves and these halves are connected mutually parallel across the same supply, the energy dissipation rate will become:
A.1 / 4 mathrmW$1 / 4 \mathrm{W}$
B.1 / 2 mathrmW$1 / 2 \mathrm{W}$
C.2 mathrm~W$2 \mathrm{~W}$
D.4 mathrm~W$4 \mathrm{~W}$
Solution
### Related Formula
P = fracV^2R_texteq$$P = \frac{V^2}{R_{\text{eq}}}$$
### Core Logic
Initially, the power (rate of energy dissipation) is W = fracV^2R$W = \frac{V^2}{R}$.
When the wire is cut into two halves, each half will have a resistance of fracR2$\frac{R}{2}$ (since resistance is directly proportional to length).
### Step 1: Equivalent Resistance
When these two halves (each of fracR2$\frac{R}{2}$) are connected in parallel, their equivalent resistance R_texteq$R_{\text{eq}}$ is:
R_texteq = fracleft(fracR2right) times left(fracR2right)fracR2 + fracR2 = fracR4$$R_{\text{eq}} = \frac{\left(\frac{R}{2}\right) \times \left(\frac{R}{2}\right)}{\frac{R}{2} + \frac{R}{2}} = \frac{R}{4}$$
### Step 2: New Power Dissipation
The new rate of energy dissipation W'$W'$ across the same potential difference V$V$ is:
W' = fracV^2R_texteq = fracV^2fracR4 = 4 fracV^2R$$W' = \frac{V^2}{R_{\text{eq}}} = \frac{V^2}{\frac{R}{4}} = 4 \frac{V^2}{R}$$W' = 4W$W' = 4W$
### Pattern Recognition
Cutting a resistor into n$n$ equal parts and connecting them in parallel reduces the total resistance by a factor of n^2$n^2$. Consequently, for a constant voltage supply, the power dissipated increases by a factor of n^2$n^2$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
Two resistance of 100Omega$100\Omega$ and 200Omega$200\Omega$ are connected in series with a battery of 4 mathrm~V$4 \mathrm{~V}$ and negligible internal resistance. A voltmeter is used to measure voltage across 100Omega$100\Omega$ resistance, which gives reading as 1 mathrm~V$1 \mathrm{~V}$. The resistance of voltmeter must be ________ Omega$\Omega$.
Numerical Answer.Answer: 200 to 200
Solution
### Related Formula
V = I R_texteq$$V = I R_{\text{eq}}$$R_textparallel = fracR_1 R_2R_1 + R_2$$R_{\text{parallel}} = \frac{R_1 R_2}{R_1 + R_2}$$
### Core Logic
Voltmeter Resistance diagram for Q60 - JEE Main 2024 Evening
The voltmeter has some internal resistance R_v$R_v$ and is connected in parallel with the 100Omega$100\Omega$ resistor.
The equivalent resistance of this parallel combination is R_p = frac100 R_v100 + R_v$R_p = \frac{100 R_v}{100 + R_v}$.
This combination is in series with the 200Omega$200\Omega$ resistor. The total voltage applied is 4 mathrm~V$4 \mathrm{~V}$.
### Step 1: Set up Voltage Divider
The voltage across the parallel combination (the voltmeter reading) is 1 mathrm~V$1 \mathrm{~V}$.
Therefore, the voltage across the 200Omega$200\Omega$ resistor must be 4 mathrm~V - 1 mathrm~V = 3 mathrm~V$4 \mathrm{~V} - 1 \mathrm{~V} = 3 \mathrm{~V}$.
Using the voltage divider rule (or equating currents since they are in series):
I = fracV_200200 = frac3200 mathrm~A$$I = \frac{V_{200}}{200} = \frac{3}{200} \mathrm{~A}$$
### Step 2: Solve for Voltmeter Resistance
The current I$I$ also flows through the parallel combination R_p$R_p$:
V_p = I R_p$V_p = I R_p$1 = left(frac3200right) left( frac100 R_v100 + R_v right)$$1 = \left(\frac{3}{200}\right) \left( \frac{100 R_v}{100 + R_v} \right)$$200 (100 + R_v) = 300 R_v$$200 (100 + R_v) = 300 R_v$$20000 + 200 R_v = 300 R_v$$20000 + 200 R_v = 300 R_v$$100 R_v = 20000$100 R_v = 20000$R_v = 200 Omega$$R_v = 200 \Omega$$
### Pattern Recognition
If the voltage splits as 1mathrmV$1\mathrm{V}$ to 3mathrmV$3\mathrm{V}$, the resistances must be in a 1:3$1:3$ ratio. So R_p = 200 / 3$R_p = 200 / 3$. Equating 100 R_v / (100+R_v) = 200/3$100 R_v / (100+R_v) = 200/3$ instantly gives R_v = 200$R_v = 200$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
More Current Electricity Questions — jee_main_2026_21_jan_morning
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