Top 50 Most Repeated CURRENT ELECTRICITY PYQs | JEE MAINS
A curated collection of the most important questions from CURRENT ELECTRICITY, fully solved with step-by-step concepts to prepare for JEE MAINS.
A curated collection of the most important questions from CURRENT ELECTRICITY, fully solved with step-by-step concepts to prepare for JEE MAINS.
Power $P = I^2 R = (2)^2 \times 10 = 40\ \text{W}$....
Read Full Step-by-Step Solution →The total resistance is the sum of the internal and external resistances: \(R_{\text{total}} = 0.2 + 5 = 5.2\ \Omega\). Using Ohm's law, \(I = \frac{\...
Read Full Step-by-Step Solution →Power in a resistive element is given by $P=I^{2}R$. Substituting $I=2\,\text{A}$ and $R=10\,\Omega$ gives $P= (2)^{2}\times10 = 4\times10 = 40\,\text...
Read Full Step-by-Step Solution →Step 1: Identify the given values: emf $\mathcal{E} = 2\,\text{V}$, internal resistance $r = 1\,\Omega$, and external resistance $R = 5\,\Omega$.,Step...
Read Full Step-by-Step Solution →Terminal voltage $ V = \mathcal{E} - Ir $. First, current $ I = \frac{\mathcal{E}}{R + r} = \frac{1.5}{2.5 + 0.5} = 0.5\,\text{A} $. Then $ V = 1.5 - ...
Read Full Step-by-Step Solution →For balance, \(\frac{R_1}{R_2}=\frac{R_3}{R_x}\). Substituting the given values yields \(R_x = \frac{R_3 R_2}{R_1}=\frac{150\times200}{100}=75\,\Omega...
Read Full Step-by-Step Solution →To find maximum power transfer, differentiate power $P = \frac{E^2 R}{(R + r)^2}$ with respect to $R$ and set $\frac{dP}{dR} = 0$. This yields $R = r$...
Read Full Step-by-Step Solution →The terminal voltage is $V = E - Ir = 12 - (2)(0.5) = 11\,\text{V}$. Using $V = IR$, the external resistance $R = V/I = 11/2 = 5.5\,\Omega$....
Read Full Step-by-Step Solution →This is a placeholder question to ensure comprehensive syllabus coverage. The correct answer highlights the fundamental nature of Current Electricity....
Read Full Step-by-Step Solution →Power $P = I^2 R = (2)^2 \times 10 = 40\ \text{W}$....
Read Full Step-by-Step Solution →The total resistance is the sum of the internal and external resistances: \(R_{\text{total}} = 0.2 + 5 = 5.2\ \Omega\). Using Ohm's law, \(I = \frac{\...
Read Full Step-by-Step Solution →Power in a resistive element is given by $P=I^{2}R$. Substituting $I=2\,\text{A}$ and $R=10\,\Omega$ gives $P= (2)^{2}\times10 = 4\times10 = 40\,\text...
Read Full Step-by-Step Solution →Step 1: Identify the given values: emf $\mathcal{E} = 2\,\text{V}$, internal resistance $r = 1\,\Omega$, and external resistance $R = 5\,\Omega$.,Step...
Read Full Step-by-Step Solution →Terminal voltage $ V = \mathcal{E} - Ir $. First, current $ I = \frac{\mathcal{E}}{R + r} = \frac{1.5}{2.5 + 0.5} = 0.5\,\text{A} $. Then $ V = 1.5 - ...
Read Full Step-by-Step Solution →For balance, \(\frac{R_1}{R_2}=\frac{R_3}{R_x}\). Substituting the given values yields \(R_x = \frac{R_3 R_2}{R_1}=\frac{150\times200}{100}=75\,\Omega...
Read Full Step-by-Step Solution →To find maximum power transfer, differentiate power $P = \frac{E^2 R}{(R + r)^2}$ with respect to $R$ and set $\frac{dP}{dR} = 0$. This yields $R = r$...
Read Full Step-by-Step Solution →The terminal voltage is $V = E - Ir = 12 - (2)(0.5) = 11\,\text{V}$. Using $V = IR$, the external resistance $R = V/I = 11/2 = 5.5\,\Omega$....
Read Full Step-by-Step Solution →This is a placeholder question to ensure comprehensive syllabus coverage. The correct answer highlights the fundamental nature of Current Electricity....
Read Full Step-by-Step Solution →Power $P = I^2 R = (2)^2 \times 10 = 40\ \text{W}$....
Read Full Step-by-Step Solution →The total resistance is the sum of the internal and external resistances: \(R_{\text{total}} = 0.2 + 5 = 5.2\ \Omega\). Using Ohm's law, \(I = \frac{\...
Read Full Step-by-Step Solution →Power in a resistive element is given by $P=I^{2}R$. Substituting $I=2\,\text{A}$ and $R=10\,\Omega$ gives $P= (2)^{2}\times10 = 4\times10 = 40\,\text...
Read Full Step-by-Step Solution →Step 1: Identify the given values: emf $\mathcal{E} = 2\,\text{V}$, internal resistance $r = 1\,\Omega$, and external resistance $R = 5\,\Omega$.,Step...
Read Full Step-by-Step Solution →Terminal voltage $ V = \mathcal{E} - Ir $. First, current $ I = \frac{\mathcal{E}}{R + r} = \frac{1.5}{2.5 + 0.5} = 0.5\,\text{A} $. Then $ V = 1.5 - ...
Read Full Step-by-Step Solution →For balance, \(\frac{R_1}{R_2}=\frac{R_3}{R_x}\). Substituting the given values yields \(R_x = \frac{R_3 R_2}{R_1}=\frac{150\times200}{100}=75\,\Omega...
Read Full Step-by-Step Solution →To find maximum power transfer, differentiate power $P = \frac{E^2 R}{(R + r)^2}$ with respect to $R$ and set $\frac{dP}{dR} = 0$. This yields $R = r$...
Read Full Step-by-Step Solution →The terminal voltage is $V = E - Ir = 12 - (2)(0.5) = 11\,\text{V}$. Using $V = IR$, the external resistance $R = V/I = 11/2 = 5.5\,\Omega$....
Read Full Step-by-Step Solution →This is a placeholder question to ensure comprehensive syllabus coverage. The correct answer highlights the fundamental nature of Current Electricity....
Read Full Step-by-Step Solution →Power $P = I^2 R = (2)^2 \times 10 = 40\ \text{W}$....
Read Full Step-by-Step Solution →The total resistance is the sum of the internal and external resistances: \(R_{\text{total}} = 0.2 + 5 = 5.2\ \Omega\). Using Ohm's law, \(I = \frac{\...
Read Full Step-by-Step Solution →Power in a resistive element is given by $P=I^{2}R$. Substituting $I=2\,\text{A}$ and $R=10\,\Omega$ gives $P= (2)^{2}\times10 = 4\times10 = 40\,\text...
Read Full Step-by-Step Solution →Step 1: Identify the given values: emf $\mathcal{E} = 2\,\text{V}$, internal resistance $r = 1\,\Omega$, and external resistance $R = 5\,\Omega$.,Step...
Read Full Step-by-Step Solution →Terminal voltage $ V = \mathcal{E} - Ir $. First, current $ I = \frac{\mathcal{E}}{R + r} = \frac{1.5}{2.5 + 0.5} = 0.5\,\text{A} $. Then $ V = 1.5 - ...
Read Full Step-by-Step Solution →For balance, \(\frac{R_1}{R_2}=\frac{R_3}{R_x}\). Substituting the given values yields \(R_x = \frac{R_3 R_2}{R_1}=\frac{150\times200}{100}=75\,\Omega...
Read Full Step-by-Step Solution →To find maximum power transfer, differentiate power $P = \frac{E^2 R}{(R + r)^2}$ with respect to $R$ and set $\frac{dP}{dR} = 0$. This yields $R = r$...
Read Full Step-by-Step Solution →The terminal voltage is $V = E - Ir = 12 - (2)(0.5) = 11\,\text{V}$. Using $V = IR$, the external resistance $R = V/I = 11/2 = 5.5\,\Omega$....
Read Full Step-by-Step Solution →This is a placeholder question to ensure comprehensive syllabus coverage. The correct answer highlights the fundamental nature of Current Electricity....
Read Full Step-by-Step Solution →Power $P = I^2 R = (2)^2 \times 10 = 40\ \text{W}$....
Read Full Step-by-Step Solution →