Question

Sham made one circuit for his Physics project. He used five resistances: two 3ohm, one 122oh
one 6ohm, one 3.6ohm and a battery of 4.5 V.
The circuit diagram and questions are given above.​

Answer 1

Answer:

2.4 and 9

Explanation:

hope this will help you

Answer 2

Given: 5 resistors are present. The potential of battery is 4.5V.

To find: (i) total resistance in parallel combination.

(ii) equivalent resistance of total circuit

(iii) Total current in circuit

(iv) current in 6Ω resistance

(v) potential difference across 3.6Ω.

Step-to-step explanation:

Step 1 of 5

The resistance in parallel connection is equal to

$\frac{1}{R_{P} }=\frac{1}{R_{1} } +\frac{1}{R_{2} }\\\frac{1}{R_{P} } =\frac{1}{6 } +\frac{1}{4 } \\\frac{1}{R_{P} } =\frac{6+4}{6*4 } \\\frac{1}{R_{P} } =\frac{10}{24 } \\\\R_{P}=2.4ohm$

Step 2 of 5

Equivalent resistance will be:

$R_{eq} =3.6+2.4+3\\R_{eq} =9ohm$

Step 3 of 5

Total current will be:

$I_{1} =\frac{R_{T*}I_{T} }{R_{1} } \\I_{1} =\frac{9*0.5}{6} \\I_{1} =0.75A$

Step 4 of 5

Current through 6Ω resistance is:

$V=IR\\I=\frac{V}{R} \\I=\frac{4.5}{6} \\I=0.75A$

Step 5 of 5

Potential difference across 3.6Ω resistance will be:

4.5V of potential difference will be across 3.6Ω.

(i) 2.4ohm

(ii) 9ohm

(iii) 0.5A

(iv) 0.75A

(v) 4.5V