The purpose of cells connected in parallel is to
Parallel cells increase current capacity, same voltage, reduce internal resistance.
Ref: NCERT Physics Class 12 Chapter 3 Current Electricity
#voltage
17 public questions tagged with this topic.
Filter questions
A 6 V battery is connected across a resistance R. The current through R is 0.4 mA. What will be the value of R?
R=V/I=6/0.4e-3=15kΩ per Ohm's law.
Ref: NCERT Physics Class 12 Chapter 3 Current Electricity
If a capacitor stores 0.12 C at 10 V, then its capacitance is
C=Q/V=0.12/10=0.012 F=12 mF.
Ref: NCERT Physics Class 12 Chapter 1 Electric Charges and Fields
In an ideal AC circuit with only a capacitor, what is the relationship between the rates of change of voltage and curren
**At resonance** V_L = I X_L = I X_C = V_C, may be larger than source voltage Q times, Q-factor = ω₀ L/R =1/(ω₀ C R)= V_L/V = V_C/V, measures sharpness, higher Q sharper resonance, bandwidth Δω = R/L = ω₀/Q, resonant frequency independent of R. In a capacitor, I = C (dV/dt) , meaning the current is directly proportional to the rate of change of voltage. Conversely, the rate of change of current relates to the second derivative of voltage, but the primary relationship is that current depends on (dV/dt) . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L -
Ref: NCERT > Physics Book > Alternating Currents > Resonance in LCR Circuit and Q-Factor
In an AC circuit with a resistor and capacitor in series, what happens to the total voltage across the components compar
**At resonance** V_L = I X_L = I X_C = V_C, may be larger than source voltage Q times, Q-factor = ω₀ L/R =1/(ω₀ C R)= V_L/V = V_C/V, measures sharpness, higher Q sharper resonance, bandwidth Δω = R/L = ω₀/Q, resonant frequency independent of R. In an RC series circuit, the voltages across the resistor ( V_R ) and capacitor ( V_C ) are 90° out of phase. The total source voltage is the vector sum, V = √(V_R² + V_C²) , which equals the applied voltage, not the algebraic sum, due to the phase difference. Applying X_L = ωL, X_C = 1/ωC, Z
Ref: NCERT > Physics Book > Alternating Currents > Resonance in LCR Circuit and Q-Factor
A \( 30 \, \mu\text{F} \) capacitor is connected to a \( 220 \, \text{V} \), \( 50 \, \text{Hz} \) AC source. What is th
**Inductive reactance** X_L = ω L =2π f L (Ω), L inductance (H), f frequency (Hz), ω=2πf angular frequency (rad/s), opposes AC, impedance Z = X_L for pure L, current lags voltage by 90°, I_rms = V_rms/X_L, I_peak = V_peak/X_L. For 85 mH, 50 Hz, X_L=2π×50×0.085=26.7 Ω, V_rms=230 V, I_rms=8.61 A, I_peak=12.18 A. X_C = (1/ω C) , ω = 2π × 50 = 314 rad/s . C = 30 × 10⁻⁶ F . X_C = (1/314 × 30 × 10⁻⁶) ≈ 106.1 Ω . RMS current: I = (V/X_C) = (220/106.1) ≈ 2.074 A . Peak current: i_m = √(2) I = 1.414 ×
Ref: NCERT > Physics Book > Alternating Currents > AC Through Inductor - Inductive Reactance
In an AC circuit with only a resistor, how does the current behave relative to the applied voltage?
**Current relative to voltage** in resistor in phase, φ=0°, power factor cos φ=1, maximum power, unlike inductor/capacitor where average power zero due to 90° phase shift, explaining why resistor heats while pure L/C does not. In a purely resistive AC circuit, the current and voltage oscillate in phase, meaning they reach their peak, zero, and minimum values simultaneously. This occurs because a resistor does not introduce any phase shift between voltage and current. Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms = V_rms/Z, P = V_rms I_rms cosφ, V_s/V_p = N_s/N_p, calculation g
Ref: NCERT > Physics Book > Alternating Currents > AC Through Resistor - Phasor and Power
A \( 5 \, \mu\text{F} \) capacitor is charged to \( 200 \, \text{V} \). What is the energy stored in it?
**Work done charging capacitor** is integral ∫ V dQ = ∫ Q/C dQ = Q²/2C, stored as electrostatic energy. When capacitor discharges, energy released as heat or work, explaining spark when shorted, energy proportional to V². U = (1/2) C V² = (1/2) × 5 × 10⁻⁶ × (200)² . U = (1/2) × 5 × 10⁻⁶ × 40000 = 0.1 J . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq = ΣC parallel, U = ½ C V² and common potential V = Q_total/C_total, result 0.1 J follows, reflecting potential-capacitance relations.
Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Energy Stored in Capacitor and Energy Density
A \( 1 \, \mu\text{F} \) capacitor is charged to \( 400 \, \text{V} \). What is the energy stored in it?
**Energy stored in capacitor** U = ½ C V² = ½ Q V = Q²/(2C) (J), C capacitance (F), V voltage (V), Q charge (C). For 4 μF charged to 250 V, U=0.5×4×10⁻⁶×62500=0.125 J. Energy resides in electric field, energy density u = ½ ε₀ E² (J/m³), E field between plates. U = (1/2) C V² = (1/2) × 1 × 10⁻⁶ × (400)² . U = (1/2) × 1 × 10⁻⁶ × 16 × 10⁴ = 0.08 J . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq = ΣC parallel, U = ½ C V²
Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Energy Stored in Capacitor and Energy Density
A \( 3 \, \mu\text{F} \) capacitor is charged to \( 300 \, \text{V} \). What is the energy stored in it?
**Work done charging capacitor** is integral ∫ V dQ = ∫ Q/C dQ = Q²/2C, stored as electrostatic energy. When capacitor discharges, energy released as heat or work, explaining spark when shorted, energy proportional to V². U = (1/2) C V² = (1/2) × 3 × 10⁻⁶ × (300)² = (1/2) × 3 × 10⁻⁶ × 9 × 10⁴ = 0.135 J . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq = ΣC parallel, U = ½ C V² and common potential V = Q_total/C_total, result 0.135 J follows, reflecting potential-capacitance relations.
Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Energy Stored in Capacitor and Energy Density
A \( 10 \, \mu\text{F} \) capacitor charged to \( 40 \, \text{V} \) is connected to an uncharged \( 10 \, \mu\text{F} \)
**Sharing of charges** when charged capacitor C₁ at V₁ connected to uncharged C₂, total charge Q = C₁ V₁ conserved, common potential V_common = Q/(C₁+C₂) = C₁ V₁/(C₁+C₂), final charges Q₁' = C₁ V_common, Q₂' = C₂ V_common. For 4 μF at 100 V (Q=4×10⁻⁴ C) connected to 4 μF uncharged, V_common=4×10⁻⁴/8×10⁻⁶=50 V. Initial charge: Q = 10 × 10⁻⁶ × 40 = 4 × 10⁻⁴ C . Total capacitance: 10 + 10 = 20 μF . Final voltage: V = (Q/C) = (4 × 10⁻⁴/20 × 10⁻⁶) = 20 V . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C =
Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Sharing of Charges, Common Potential and Applications
A \( 7 \, \mu\text{F} \) capacitor is charged to \( 400 \, \text{V} \). What is the energy stored in it?
**Work done charging capacitor** is integral ∫ V dQ = ∫ Q/C dQ = Q²/2C, stored as electrostatic energy. When capacitor discharges, energy released as heat or work, explaining spark when shorted, energy proportional to V². U = (1/2) C V² = (1/2) × 7 × 10⁻⁶ × (400)² . U = (1/2) × 7 × 10⁻⁶ × 160000 = 0.56 J . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq = ΣC parallel, U = ½ C V² and common potential V = Q_total/C_total, result 0.56 J follows, reflecting potential-capacitance relations.
Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Energy Stored in Capacitor and Energy Density
