Three resisters 5000 500 50 ohm are connected in series with the 555 voltage main what is the current About the author Jade
Answer: R1= 5000ohms R2= 500ohms R3= 50ohms Rresultant = R1+R2+R3 R resultant = 5000+500+50 R resultant = 5550ohms v = 555v I = V\R I= 555\ 5550 I = 1\10 AMPERE Explanation: Reply
Answer: 0.1 A Explanation: As per the provided information in the given question, we have : Three resistors of 5000Ω, 500Ω and 50Ω are connected in series. Voltage or potential difference is 550 V. We are asked to calculate how much current is flowing in the circuit. In order to calculate how much current is flowing in the circuit, we need to find the equivalent resistance in the circuit. Since the resistors are connected in series combination, we need to apply the formula of equivalent resistance in series combination. As we know that, when resistors are connected in series combination, then equivalent resistance pf the circuit is given by, [tex] \twoheadrightarrow \quad \sf { R_S = R_1 + R_2 + R_3 + \dots R_n} [/tex] Here, [tex] \rm R_1 [/tex] = 5000Ω [tex] \rm R_2 [/tex] = 500Ω [tex] \rm R_3 [/tex] = 50Ω Substituting the values, we get: [tex] \twoheadrightarrow \quad \sf { R_S = R_1 + R_2 + R_3} [/tex] [tex] \twoheadrightarrow \quad \sf { R_S = (5000 + 500 +50) \; \Omega} [/tex] [tex] \twoheadrightarrow \quad \bf \underline { R_S = 5550 \; \Omega} [/tex] Now, let’s calculate the current by using ohm’s law. We know that, according to Ohm’s law : [tex] \twoheadrightarrow \quad \sf { V = IR} [/tex] V denotes potential difference I denotes current R denotes resistance Substituting values, [tex] \twoheadrightarrow \quad \sf { 555 = I \times 5550} [/tex] [tex] \twoheadrightarrow \quad \sf { \cancel{\dfrac{555}{5550}} \; A= I } [/tex] [tex] \twoheadrightarrow \quad \sf { \dfrac{1}{10} \; A= I } [/tex] [tex] \twoheadrightarrow \quad \bf \underline{ 0.1 \; A= I } [/tex] Therefore, 0.1 Ampere of current is flowing in the circuit. Reply
Answer:
R1= 5000ohms R2= 500ohms R3= 50ohms
Rresultant = R1+R2+R3
R resultant = 5000+500+50
R resultant = 5550ohms
v = 555v
I = V\R
I= 555\ 5550
I = 1\10 AMPERE
Explanation:
Answer:
0.1 A
Explanation:
As per the provided information in the given question, we have :
We are asked to calculate how much current is flowing in the circuit.
In order to calculate how much current is flowing in the circuit, we need to find the equivalent resistance in the circuit.
Since the resistors are connected in series combination, we need to apply the formula of equivalent resistance in series combination.
As we know that, when resistors are connected in series combination, then equivalent resistance pf the circuit is given by,
[tex] \twoheadrightarrow \quad \sf { R_S = R_1 + R_2 + R_3 + \dots R_n} [/tex]
Here,
Substituting the values, we get:
[tex] \twoheadrightarrow \quad \sf { R_S = R_1 + R_2 + R_3} [/tex]
[tex] \twoheadrightarrow \quad \sf { R_S = (5000 + 500 +50) \; \Omega} [/tex]
[tex] \twoheadrightarrow \quad \bf \underline { R_S = 5550 \; \Omega} [/tex]
Now, let’s calculate the current by using ohm’s law. We know that, according to Ohm’s law :
[tex] \twoheadrightarrow \quad \sf { V = IR} [/tex]
Substituting values,
[tex] \twoheadrightarrow \quad \sf { 555 = I \times 5550} [/tex]
[tex] \twoheadrightarrow \quad \sf { \cancel{\dfrac{555}{5550}} \; A= I } [/tex]
[tex] \twoheadrightarrow \quad \sf { \dfrac{1}{10} \; A= I } [/tex]
[tex] \twoheadrightarrow \quad \bf \underline{ 0.1 \; A= I } [/tex]
Therefore, 0.1 Ampere of current is flowing in the circuit.