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dimaraw [331]
3 years ago
6

You have a 5 V power source, a 25 W resistor, a 50 W resistor, and some wire. Clearly identify the formula(s) you are using, and

circle your final answer. a) You arrange the two resistors in series with the battery, what is the equivalent resistance of the circuit? b) What is the current flowing through each resistor and the battery (for the series circuit)? c) If you arrange the two resistors in parallel with the battery, what is the equivalent resistance of the circuit? d) What is the current flowing through each resistor and the battery (for the parallel circuit)? Note that the current is not necessarily the same through the battery and each resistor, so specify the current in each individual circuit element.
Physics
1 answer:
11111nata11111 [884]3 years ago
5 0

Answer:

Explanation:

given,

Power source = 5 V

R₁ = 25 Ω

R₂ = 50 Ω

a) when they are arranged in series

R = R₁ + R₂

R = 25 + 50

R = 75 Ω

b) current = I = \dfrac{V}{R}

                   I =  \dfrac{5}{75}

                   I = 0.067 A

so, the resistors are connected in series hence current in both the resistor will be 0.067 A.

c) when they are arranged in parallel

\dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2}

\dfrac{1}{R} = \dfrac{1}{25} + \dfrac{1}{50}

R = 16.67 Ω

d) current through the battery  I = \dfrac{V}{R}

                                                   I = \dfrac{5}{16.67} = 0.3 A

current through 25 Ω resistor = \dfrac{V}{R} = \dfrac{5}{25} = 0.2 A

current through 50 Ω resistor = \dfrac{V}{R} = \dfrac{5}{50} = 0.1 A

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KE = (½) * (m) * (v²) ;  to solve for "(v)" ; velocity, which is very similar to                                          the "speed"; 
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→ We have:  (½)*(m)* (v²) = (KE)  ; we isolate, "m" (mass) on one side of the equation:
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→ We divide each side of the equation by: "[(½)* (m)]" ; 
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