Physics
Current Electricity and Circuits
377 Questions
Current electricity and circuits questions cover resistors, EMF, internal resistance, and power calculations in series and parallel configurations. Solving these builds a strong understanding of electrical principles and circuit analysis. These physics problems are highly relevant for technical and science aptitude tests.
Resistor combinationsPower dissipationEMF and internal resistanceAC circuit analysisOperational amplifiers
Current Electricity and Circuits Questions
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ohm
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Thelma z Ohms
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Georg Simon Ohm
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martin ohms
C
Correct answer
Explanation
The ohm, the SI unit of electrical resistance, is named after Georg Simon Ohm, the German physicist and mathematician who discovered Ohm's law. This fundamental law states that current through a conductor is directly proportional to voltage and inversely proportional to resistance.
A
Correct answer
Explanation
In a series circuit, the total voltage is divided across the resistors in proportion to their resistance values (Voltage Division Rule). A resistor with higher resistance will have a larger voltage drop across it. This is because the same current flows through all components in series, and V = IR for each resistor.
C
Correct answer
Explanation
For three identical resistors in parallel, the equivalent resistance is R/n where n is the number of resistors. Here: 30kΩ / 3 = 10kΩ. Series would add them (90kΩ), and the other options (300kΩ, 33kΩ) are not parallel combinations of equal resistors.
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V02 = $\sqrt2$V01
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V02 = e2 V01
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V02 = V01 In 2
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V01 - V02 = VT In 2
D
Correct answer
Explanation
Here the inverting terminal is at virtual ground and the current in resistor and diode current is equal i.e.
$$
\begin{aligned}
I_R = & \quad I_D \\
or \qquad \dfrac{V_i}{R} = & \quad I_s e^{\frac{V_0}{V_r}} \\
or \qquad V_D = &\quad V_T\ ln \dfrac{V_i}{I_sR} \\
\text{For the first condition}\\
V_D = 0 - V_{o1} = V_T \ ln \ \dfrac{2}{I_sR}\\
\text{For the first condition}\\
V_D = 0 - V_{o1} = V_T \ ln \ \dfrac{4}{I_sR}\\
\text{Subtracting the above equation}\\
V_{o1} - V_{o2} = &\quad V_T\ ln\dfrac{4}{I_sR} - V_T\ ln \dfrac{2}{I_sR} \\
or \quad V_{o1} - V_{o2} = &\quad V_T\ ln\dfrac{4}{2} = V_T\ ln 2
\end{aligned}
$$
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0.25cos $\omega t$ mV
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1cos $\omega t$ mV
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2cos $\omega t$ mV
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22cos $\omega t$ mV
B
Correct answer
Explanation
The forward resistance of each diode is
$$
\begin{aligned}
r & = \dfrac{V_T}{I_C} \dfrac{25 \ mV}{1 \ mA } = 25\Omega \\
Thus \quad V_{ac} \quad & = V_i \times \left( \dfrac{4(r)}{4(r) + 9900} \right) \\
& = 100 \ mV \ cos(\omega t)0.01 = 1\ cos(\omega t)
\ mV
\end{aligned}
$$
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2 M$\Omega$ and 2 k$\Omega$
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$2\ M\Omega \quad and \quad \dfrac{20}{11} K\Omega$
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Infinity and 2 k$\Omega$
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$Infinity \quad and \quad \dfrac{20}{11} K\Omega$
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$ - \dfrac{V_s}{R_2}$
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$ \dfrac{V_s}{R_2}$
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$- \dfrac{V_s}{R_L}$
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$ \dfrac{V_s}{R_1}$
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1.875 mS and 3.41
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1.875 mS and - 3.41
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3.3 mS and -6
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3.3 mS and 6
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5.625 mA and 8.75 V
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4.500 mA and 11.00 V
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7.500 mA and 5.00 V
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6.250 mA and 7.50 V
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– 1 + Cos ($\omega t$)
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sin ($\omega t$)
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1 – cos ($\omega t$)
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1 – sin ($\omega t$)
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258$\Omega$
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1252$\Omega$
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93 K$\Omega$
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$\infty$
B
Correct answer
Explanation
By using given solution figure and solution
Rin = 1000 + [(9300 * 259)/(9300 + 259)] = 1252 om
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$\dfrac{30}{4}K\Omega$
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10 k$\Omega$
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40 k$\Omega$
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infinite
B
Correct answer
Explanation
Since the inverting terminal is at virtual ground, the current flowing through the voltage source is
$$
I_s = \dfrac{V_s}{10K} \\
or \quad \dfrac{V_s}{I_s} = 10 \ K\Omega = R_{in}
$$
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|Av| $\approx$ 200
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|Av| $\approx$ 100
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|Av| $\approx$ 20
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|Av| $\approx$ 10
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7.00 V to 7.29 V
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7.14 V to 7.29 V
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7.14 V to 7.43 V
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7.29 V to 7.43 V
