Please explain the reasoning as well. Consider the following circuit. Rank the b
ID: 1433844 • Letter: P
Question
Please explain the reasoning as well.
Consider the following circuit. Rank the brightness of the bulbs. Explain your reasoning. If bulb B were removed from the circuit, would the current through bulb C increase, decrease or remain the same? Explain your reasoning. bulb D increase, decrease or remain the same? Explain your reasoning. the battery increase, decrease or remain the same? Explain your reasoning. If bulb B were removed from the circuit, would the potential difference across bulb C increase, decrease or remain the same? Explain your reasoning. bulb D increase, decrease or remain the same? Explain your reasoning. bulb A increase, decrease or remain the same? Explain your reasoning. the battery increase, decrease or remain the same? Explain your reasoning.Explanation / Answer
a)
D >C >A = B
the resistance of bulb D is smaller than the resistance of combination of bulbs A, B and C. Hence current flowing through branch of D is greater as compared to that of branch contaning A, B and C. the current in bulb A and B divides equally as they are in parallel and have same resistance. hence A and B have equal brightness . the current in C is sum of current in A and B , hence current in C is greater than A and B and therefore C has greater brightness as compared to A or C .
b)
i)
decrease
As bulb B is removed , combined resistance of branch containing A, B and C will increases .as a result, it decrease the current in branch containing A, B and C .since as resistance increase , current decreases to have same constant voltage.
ii)
As bulb B is removed , the voltage across the bulb D does not change and remains equal to battery voltage . hence the current in bulb D remains same
iii)
The current in battery is combination of current in bulb D and Current in branch containing A, B and C. Since the current in D remains same and current in branch containing A,B,C decreases, hence current in battery decreases too.
c)
i)
the current in C decreases as the bulB B is removed . voltage is given as
V = iR
as i decreases , Voltage V decreases since resistance R remains constant
ii)
the bulb D is in parallel with the battery hence, the voltage across D remains same
iii)
since the current in branch containing A,B and C decreases , the current in bulb A decreases too
hence the potential difference across bulb A decrease
iv)
the battery Voltage remains same