We read your photo as:

Define the term emf of a battery.

Physics

ScanSolve AI Answer

Answer

Rext=13.375 ΩR_{ext} = \text{}13.375 \, \Omega\text{}

I=3 AI = \text{}3 \, \text{A}\text{}

41.625 V\text{}41.625 \, \text{V}\text{}

Step-by-step solution

7 steps
  1. Define EMF.

    Electromotive force (EMF) is the maximum potential difference that a battery or source of electrical energy can provide across its terminals when no current is flowing in the external circuit. It represents the total energy supplied per unit charge by the source.

  2. Explain why the voltmeter reading decreases.

    When the switch S is closed, current flows through the external circuit. This current also flows through the internal resistance (rr) of the battery, causing a potential drop (IrIr) across the internal resistance. The voltmeter measures the terminal potential difference (VterminalV_{terminal}), which is given by Vterminal=ε−IrV_{terminal} = \varepsilon - Ir. Since Ir>0Ir > 0 when current flows, the terminal potential difference is less than the EMF, causing the voltmeter reading to decrease.

  3. Calculate the total external resistance.

    Resistors R2R_2 and R3R_3 are in parallel. Their equivalent resistance (RpR_p) is:

    1Rp=1R2+1R3\frac{1}{R_p} = \frac{1}{R_2} + \frac{1}{R_3} 1Rp=125 Ω+115 Ω\frac{1}{R_p} = \frac{1}{25 \, \Omega} + \frac{1}{15 \, \Omega} 1Rp=375 Ω+575 Ω=875 Ω\frac{1}{R_p} = \frac{3}{75 \, \Omega} + \frac{5}{75 \, \Omega} = \frac{8}{75 \, \Omega} Rp=758 Ω=9.375 ΩR_p = \frac{75}{8} \, \Omega = 9.375 \, \Omega

    Resistor R1R_1 is in series with RpR_p. The total external resistance (RextR_{ext}) is:

    Rext=R1+RpR_{ext} = R_1 + R_p Rext=4 Ω+9.375 ΩR_{ext} = 4 \, \Omega + 9.375 \, \Omega Rext=13.375 ΩR_{ext} = \boxed{\text{}13.375 \, \Omega\text{}}
  4. Calculate the reading on the ammeter.

    The decrease in voltmeter reading (1.5 V1.5 \, \text{V}) is due to the potential drop across the internal resistance (IrIr).

    Ir=1.5 VIr = 1.5 \, \text{V}

    Given the internal resistance r=0.5 Ωr = 0.5 \, \Omega:

    I(0.5 Ω)=1.5 VI (0.5 \, \Omega) = 1.5 \, \text{V} I=1.5 V0.5 ΩI = \frac{1.5 \, \text{V}}{0.5 \, \Omega} I=3 AI = \boxed{\text{}3 \, \text{A}\text{}}
  5. Calculate the EMF of the battery.

    The EMF (ε\varepsilon) is given by the formula ε=I(Rext+r)\varepsilon = I(R_{ext} + r).

    ε=3 A(13.375 Ω+0.5 Ω)\varepsilon = 3 \, \text{A} (13.375 \, \Omega + 0.5 \, \Omega) ε=3 A(13.875 Ω)\varepsilon = 3 \, \text{A} (13.875 \, \Omega) ε=41.625 V\varepsilon = \boxed{\text{}41.625 \, \text{V}\text{}}
  6. Evaluate the statement about current through R2R_2 and R3R_3.

    The statement is: "The current through resistor R3R_3 is larger than the current through resistor R2R_2."
    Resistors R2R_2 (25 Ω25 \, \Omega) and R3R_3 (15 Ω15 \, \Omega) are connected in parallel, meaning the voltage across them is the same. According to Ohm's Law (I=V/RI = V/R), current is inversely proportional to resistance. Since R3<R2R_3 < R_2, the current through R3R_3 will be greater than the current through R2R_2.
    Therefore, the statement is CORRECT.

  7. Determine the effect of removing R1R_1 on the EMF.

    The EMF of a battery is an intrinsic property of the battery itself, determined by its chemical composition and internal structure. It represents the maximum potential difference the battery can provide. Removing an external resistor changes the total external resistance of the circuit, which affects the total current and the terminal potential difference, but it does not change the fundamental EMF of the battery.
    Therefore, the EMF of the battery REMAINS THE SAME.

Was this helpful?

Try one yourself

A new problem that tests the same skill.

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question
Handwritten step-by-step solution preview

Handwritten Step-by-Step Solution

Get this as a handwritten image on WhatsApp — like a tutor wrote it for you.

Try on WhatsApp