Consider the circuit shown in Fig. P 25.83 . The emf source has negligible internal resistance.

step 1 Okay, so the EMF in the circuits can be equal to v1 plus v2. V1 is the potential difference acro

Consider the circuit shown in Fig. P 25.83 . The emf source...

Consider the circuit shown in Fig. $\mathrm{P} 25.83 .$ The emf source has negligible internal resistance. The resistors have resistances $R_{1}=6.00 \Omega$ and $R_{2}=4.00 \Omega .$ The capacitor has capacitance $C=9.00 \mu \mathrm{F}$ When the capacitor is fully charged, the magnitude of the charge on its plates is $Q=36.0 \mu \mathrm{C} .$ Calculate the emf $\mathcal{E} .$

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Key Concepts

Capacitor Charge-Voltage Relationship

This concept links the amount of electric charge stored on a capacitor to the voltage across its plates through the equation Q = CV, where Q is the charge, C is the capacitance, and V is the voltage. It is fundamental in understanding how a capacitor stores energy and how its voltage changes when different charges or capacitances are involved.

Steady-State Behavior in DC Circuits

In a DC circuit, when a capacitor is fully charged, it reaches a steady state where no current flows through the branch containing the capacitor. This behavior is crucial because it effectively turns the capacitor into an open circuit, meaning that the voltage across it remains constant and is determined solely by the charge stored and its capacitance.

Ohm’s Law and Voltage Division

Ohm’s Law, which expresses the relationship between voltage, current, and resistance (V = IR), underpins the analysis of circuits composed of resistors. In circuits where components are arranged in series, the voltage divider rule can be used to determine the voltage drop across each resistor. This principle helps relate the total emf supplied by the battery to the individual voltage drops, including the voltage across the capacitor when the circuit is at steady state.

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